A Survey on Cauchy-Buniakowsky-Schwartz Type Discrete Inequalities

A Survey on Cauchy-Buniakowsky-Schwartz Type Discrete Inequalities Sever Silvestru Dragomir School of Computer Science and Mathematics Victoria University, PO Box 14428, MCMC 8001, Melbourne, Victoria, Australia. January 20, 2003

ii

Contents 1 (CBS) – Type Inequalities 1.1 (CBS) −Inequality for Real Numbers . . . . . . . 1.2 (CBS) −Inequality for Complex Numbers . . . . 1.3 An Additive Generalisation . . . . . . . . . . . . 1.4 A Related Additive Inequality . . . . . . . . . . . 1.5 A Parameter Additive Inequality . . . . . . . . . 1.6 A Generalisation Provided by Young’s Inequality 1.7 Further Generalisations via Young’s Inequality . . 1.8 A Generalisation Involving J−Convex Functions . 1.9 A Functional Generalisation . . . . . . . . . . . . 1.10 A Generalisation for Power Series . . . . . . . . . 1.11 A Generalisation of Callebaut’s Inequality . . . . 1.12 Wagner’s Inequality for Real Numbers . . . . . . 1.13 Wagner’s inequality for Complex Numbers . . . .

. . . . . . . . . . . . .

2 Refinements of the (CBS) −Inequality 2.1 A Refinement in Terms of Moduli . . . . . . . . . . 2.2 A Refinement for a Sequence Whose Norm is One . 2.3 A Second Refinement in Terms of Moduli . . . . . . 2.4 A Refinement for a Sequence Less than the Weights 2.5 A Conditional Inequality Providing a Refinement . 2.6 A Refinement for Non-Constant Sequences . . . . . 2.7 De Bruijn’s Inequality . . . . . . . . . . . . . . . . 2.8 McLaughlin’s Inequality . . . . . . . . . . . . . . . 2.9 A Refinement due to Daykin-Eliezer-Carlitz . . . . 2.10 A Refinement via Dunkl-Williams’ Inequality . . . 2.11 Some Refinements due to Alzer and Zheng . . . . . iii

. . . . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . . . .

1 1 2 4 7 8 10 12 18 20 23 25 27 29

. . . . . . . . . . .

35 35 38 41 44 47 50 55 57 58 60 61

iv 3 Functional Properties 3.1 A Monotonicity Property . . . . . . . . . . . . . 3.2 A Superadditivity Property in Terms of Weights 3.3 The Superadditivity as an Index Set Mapping . 3.4 Strong Superadditivity in Terms of Weights . . 3.5 Strong Superadditivity as an Index Set Mapping 3.6 Another Superadditivity Property . . . . . . . . 3.7 The Case of Index Set Mapping . . . . . . . . . 3.8 Supermultiplicity in Terms of Weights . . . . . 3.9 Supermultiplicity as an Index Set Mapping . . .

CONTENTS

. . . . . . . . .

4 Counterpart Inequalities 4.1 The Cassels’ Inequality . . . . . . . . . . . . . . . 4.2 The P´olya-Szeg¨o Inequality . . . . . . . . . . . . 4.3 The Greub-Rheinboldt Inequality . . . . . . . . . 4.4 A Cassels’ Type Inequality for Complex Numbers 4.5 A Counterpart Inequality for Real Numbers . . . 4.6 A Counterpart Inequality for Complex Numbers . 4.7 Shisha-Mond Type Inequalities . . . . . . . . . . 4.8 Zagier Type Inequalities . . . . . . . . . . . . . . 4.9 A Counterpart in Terms of the sup −Norm . . . . 4.10 A Counterpart in Terms of the 1−Norm . . . . . 4.11 A Counterpart in Terms of the p−Norm . . . . . 4.12 A Counterpart Via an Andrica-Badea Result . . . 4.13 A Refinement of Cassels’ Inequality . . . . . . . . 4.14 Two Counterparts Via Diaz-Metcalf Results . . . ˇ 4.15 Some Counterparts Via the Cebyˇ sev Functional . 4.16 Another Counterpart via a Gr¨ uss Type Result . .

. . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . . . . . . . . .

71 71 73 75 78 80 83 86 89 93

. . . . . . . . .

. . . . . . . . .

. . . . . . . . . . . . . . . .

101 . 101 . 104 . 106 . 107 . 110 . 113 . 116 . 118 . 121 . 124 . 127 . 130 . 133 . 137 . 140 . 147

5 Related Inequalities 155 5.1 Ostrowski’s Inequality for Real Sequences . . . . . . . . . . . . 155 5.2 Ostrowski’s Inequality for Complex Sequences . . . . . . . . . 156 5.3 Another Ostrowski’s Inequality . . . . . . . . . . . . . . . . . 159 5.4 Fan and Todd Inequalities . . . . . . . . . . . . . . . . . . . . 161 5.5 Some Results for Asynchronous Sequences . . . . . . . . . . . 163 5.6 An Inequality via A − G − H Mean Inequality . . . . . . . . . 164 5.7 A Related Result via Jensen’s Inequality for Power Functions . 166 5.8 Inequalities Derived from the Double Sums Case . . . . . . . . 167

CONTENTS 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21

A Functional Generalisation for Double Sums . . . . A (CBS) −Type Result for Lipschitzian Functions . . An Inequality via Jensen’s Discrete Inequality . . . . An Inequality via Lah-Ribari´c Inequality . . . . . . . An Inequality via Dragomir-Ionescu Inequality . . . . An Inequality via a Refinement of Jensen’s Inequality Another Refinement via Jensen’s Inequality . . . . . An Inequality via Slater’s Result . . . . . . . . . . . An Inequality via an Andrica-Ra¸sa Result . . . . . . An Inequality via Jensen’s Result for Double Sums . ˇ Some Inequalities for the Cebyˇ sev Functional . . . . . ˇ Other Inequalities for the Cebyˇ sev Functional . . . . ˇ Bounds for the Cebyˇsev Functional . . . . . . . . . .

v . . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

169 171 173 174 176 178 182 185 187 190 192 195 197

vi

CONTENTS

Preface The Cauchy-Buniakowski-Schwartz inequality, or for short, the (CBS)− inequality, plays an important role in different branches of Modern Mathematics including Hilbert Spaces Theory, Probability & Statistics, Classical Real and Complex Analysis, Numerical Analysis, Qualitative Theory of Differential Equations and their applications. The main purpose of this survey is to identify and highlight the discrete inequalities that are connected with (CBS)− inequality and provide refinements, counterparts and reverse results as well as to study some functional properties of certain mappings that can be naturally associated with this inequality such as superadditivity, supermultiplicity, the strong versions of these and the corresponding monotonicity properties. Many companions and related results both for real and complex numbers are also presented. The first chapter is devoted to a number of (CBS)− type inequalities that provides not only natural generalizations but also several extensions for different classes of analytic functions of a real variable. A generalization of the Wagner inequality for complex numbers is obtained. Several results discovered by the author in the late eighties and published in different journals of lesser circulation are also surveyed. The second chapter contains different refinements of the (CBS)− inequality including de Bruijn’s inequality, McLaughlin’s inequality, the DaykinEliezer-Carlitz result in the version presented by Mitrinovi´c-Peˇcari´c and Fink as well as the refinements of a particular version obtained by Alzer and Zheng. A number of new results obtained by the author, which are connected with the above ones, are also presented. Chapter 3 is devoted to the study of functional properties of different mappings naturally associated to the (CBS)− inequality. Properties such as superadditivity, strong superadditivity, monotonicity and supermultiplicity and the corresponding inequalities are mentioned. vii

viii

CONTENTS

In the next chapter, Chapter 4, counterpart results for the (CBS)− inequality are surveyed. The results of Cassels, P´olya-Szeg¨o, Greub-Rheinbold, Shisha-Mond and Zagier are presented with their original proofs. New results and versions for complex numbers are also obtained. Counterparts in terms of p−norms of the forward difference recently discovered by the author and some refinements of Cassels and P´olya-Szeg¨o results obtained via AndricaBadea inequality are mentioned. Some new facts derived from Gr¨ uss type inequalities are also pointed out. Chapter 5 is devoted to various inequalities related to the (CBS)− inequality. The two inequalities obtained by Ostrowski and Fan-Todd results are presented. New inequalities obtained via Jensen type inequality for conˇ vex functions are derived, some inequalities for the Ceby¸ sev functionals are pointed out. Versions for complex numbers that generalize Ostrowski results are also emphasised. It was one of the main aims of the survey to provide complete proofs for the results considered. We also note that in most cases only the original references are mentioned. Being self contained, the survey may be used by both postgraduate students and researchers interested in Theory of Inequalities & Applications as well as by Mathematicians and other Scientists dealing with numerical computations, bounds and estimates where the (CBS)− inequality may be used as a powerful tool. The author intends to continue this survey with another one devoted to the functional and integral versions of the (CBS)− inequality. The corresponding results holding in inner-product and normed spaces will be considered as well.

The Author,

December 2002. Melbourne, Australia.

Chapter 1 (CBS) – Type Inequalities 1.1

(CBS) −Inequality for Real Numbers

The following inequality is known in the literature as Cauchy’s or CauchySchwartz’s or Cauchy-Buniakowski-Schwartz’s inequality. For simplicity, we shall refer to it throughout this work as the (CBS) −inequality. ¯ = (b1 , . . . , bn ) are sequences of real Theorem 1 If ¯ a = (a1 , . . . , an ) and b numbers, then !2 n n n X X X 2 ak b k ≤ (1.1) ak b2k k=1

k=1

k=1

¯ are proportional, i.e., there with equality if and only if the sequences ¯ a and b is a r ∈ R such that ak = rbk for each k ∈ {1, . . . , n} . Proof. 1. Consider the quadratic polynomial P : R → R, P (t) =

n X

(ak t − bk )2 .

(1.2)

k=1

It is obvious that P (t) =

n X

! a2k

2

t −2

k=1

n X k=1

1

! ak b k

t+

n X k=1

b2k

2

CHAPTER 1. (CBS) – TYPE INEQUALITIES for any t ∈ R. Since P (t) ≥ 0 for any t ∈ R it follows that the discriminant ∆ of P is negative, i.e., !2 n n n X X X 1 0≥ ∆= ak b k − a2k b2k 4 k=1 k=1 k=1 and the inequality (1.1) is proved. 2. If we use Lagrange’s identity n X i=1

a2i

n X

b2i

−

i=1

n X

!2 ai b i

i=1

n 1X = (ai bj − aj bi )2 2 i,j=1 X = (ai bj − aj bi )2

(1.3)

1≤i<j≤n

then (1.1) obviously holds. The equality holds in (1.1) iff (ai bj − aj bi )2 = 0 ¯ are for any i, j ∈ {1, . . . , n} which is equivalent with the fact that ¯ a and b proportional. Remark 2 The inequality (1.1) apparently was firstly mentioned in the work [2] of A.L. Cauchy in 1821. The integral form was obtained in 1859 by V. Buniakowski [1]. The corresponding version for inner-product spaces is mainly known as Schwartz’s inequality.

1.2

(CBS) −Inequality for Complex Numbers

The following version of the (CBS) −inequality for complex numbers holds [3, p. 84]. ¯ = (b1 , . . . , bn ) are sequences of comTheorem 3 If ¯ a = (a1 , . . . , an ) and b plex numbers, then 2 n n n X X X 2 ak b k ≤ |ak | |bk |2 , (1.4) k=1

k=1

k=1

1.2. (CBS) −INEQUALITY FOR COMPLEX NUMBERS

3

with equality if and only if there is a complex number c ∈ C such that ak = c¯bk for any k ∈ {1, . . . , n} . Proof. 1. For any complex number λ ∈ C one has the equality n n X X

¯ k ak − λ¯bk 2 = ak − λ¯bk a ¯k − λb k=1

=

k=1 n X

2

2

|ak | + |λ|

k=1

n X

(1.5)

¯ |bk | − 2 Re λ 2

k=1

n X

! ak b k

.

k=1

If in (1.5) we choose λ0 ∈ C, Pn ak b k λ0 := Pnk=1 2, k=1 |bk |

¯ 6= 0 b

then we get the identity Pn n n 2 X 2 X | ak b k | 2 ak − λ0¯bk = 0≤ |ak | − Pk=1 n 2 , |b | k k=1 k=1 k=1

(1.6)

which proves (1.4). By virtue of (1.6), we conclude that equality holds in (1.4) if and only if ak = λ0¯bk for any k ∈ {1, . . . , n} . 2. Using Binet-Cauchy’s identity for complex numbers n X i=1

xi yi

n X i=1

zi ti −

n X i=1

xi ti

n X

zi yi

i=1

n 1X (xi zj − xj zi ) (yi tj − yj ti ) = 2 i,j=1 X = (xi zj − xj zi ) (yi tj − yj ti ) 1≤i<j≤n

(1.7)

4

CHAPTER 1. (CBS) – TYPE INEQUALITIES for the choices xi = a ¯i , zi = bi , yi = ai , ti = ¯bi , i = {1, . . . , n}, we get 2 n n n n X X X 1X 2 2 |¯ ai b j − a ¯j bi |2 (1.8) |ai | |bi | − ai bi = 2 i,j=1 i=1 i=1 i=1 X = |¯ ai b j − a ¯j bi |2 . 1≤i<j≤n

Now the inequality (1.4) is a simple consequence of (1.8). The case of equality is obvious by the identity (1.8) as well.

Remark 4 By the (CBS) −inequality for real numbers and the generalised triangle inequality for complex numbers n n X X |zi | ≥ zi , zi ∈ C, i ∈ {1, . . . , n} i=1

i=1

we also have 2 n X a b k k ≤ k=1

n X

!2 |ak bk |

≤

n X

|ak |2

k=1

k=1

n X

|bk |2 .

k=1

Remark 5 The Lagrange identity for complex numbers stated in [3, p. 85] is wrong. It should be corrected as in (1.8).

1.3

An Additive Generalisation

The following generalisation of the (CBS) −inequality was obtained in [4, p. 5]. ¯ = (b1 , . . . , bn ) , ¯ Theorem 6 If ¯ a = (a1 , . . . , an ) , b c = (c1 , . . . , cn ) and ¯ d = (d1 , . . . , dn ) are sequences of real numbers and p ¯ = (p1 , . . . , pn ) , q ¯ = (q1 , . . . , qn ) are nonnegative, then n X i=1

pi a2i

n X i=1

qi b2i

+

n X i=1

pi c2i

n X i=1

qi d2i

≥2

n X i=1

p i ai c i

n X

qi bi di .

(1.9)

i=1

If p ¯ and q ¯ are sequences of positive numbers, then the equality holds in (1.9) iff ai bj = ci dj for any i, j ∈ {1, . . . , n} .

1.3. AN ADDITIVE GENERALISATION

5

Proof. We will follow the proof from [4]. From the elementary inequality a2 + b2 ≥ 2ab for any a, b ∈ R

(1.10)

with equality iff a = b, we have a2i b2j + c2i d2j ≥ 2ai ci bj dj for any i, j ∈ {1, . . . , n} .

(1.11)

Multiplying (1.11) by pi qj ≥ 0, i, j ∈ {1, . . . , n} and summing over i and j from 1 to n, we deduce (1.9). If pi , qj > 0 (i = 1, . . . , n) , then the equality holds in (1.9) iff ai bj = ci dj for any i, j ∈ {1, . . . , n} . Remark 7 The condition ai bj = ci dj for ci 6= 0, bj 6= 0 (i, j = 1, . . . , n) is d ¯ d ¯ are proportional equivalent with acii = bjj (i, j = 1, . . . , n) , i.e., ¯ a, ¯ c and b, with the same constant k. Remark 8 If in (1.9) we choose pi = qi = 1 (i = 1, . . . , n) , ci = bi , and di = ai (i = 1, . . . , n) , then we recapture the (CBS) −inequality. The following corollary holds [4, p. 6]. ¯ ¯ ¯ are nonnegative, then Corollary 9 If ¯ a, b, c and d " n # n n n n n X X X X 1 X 3 X 3 3 3 2 2 a ci b di + c i ai di bi ≥ ai c i b2i d2i , 2 i=1 i i=1 i i=1 i=1 i=1 i=1 " n # n n n X X X 1 X 2 a bi di · b2i ai ci + c2i bi di · d2i ai ci ≥ 2 i=1 i i=1 i=1 i=1

n X

(1.12)

!2 ai bi ci di

. (1.13)

i=1

Another result is embodied in the following corollary [4, p. 6]. ¯ ¯ ¯ are sequences of positive and real numbers, Corollary 10 If ¯ a, b, c and d then: " n # n n n n n X X X 1 X a3i X b3i X + ai c i bi di ≥ a2i b2i , (1.14) 2 i=1 ci i=1 di i=1 i=1 i=1 i=1 " n # !2 n n n n X X 1 X a2i bi X b2i ai X + bi c i ai di ≥ ai b i . (1.15) 2 i=1 ci i=1 di i=1 i=1 i=1

6

CHAPTER 1. (CBS) – TYPE INEQUALITIES Finally, we also have [4, p. 6].

¯ are positive, then Corollary 11 If ¯ a, and b  n n 1 X a3i X b3i − 2 i=1 bi i=1 ai

n X

!2  ≥

ai b i

i=1

n X

a2i

i=1

n X

b2i −

i=1

n X

!2 ≥ 0.

ai b i

i=1

The following version for complex numbers also holds. ¯ = (b1 , . . . , bn ) , ¯ Theorem 12 Let ¯ a = (a1 , . . . , an ) , b c = (c1 , . . . , cn ) and ¯ d = (d1 , . . . , dn ) be sequences of complex numbers and p ¯ = (p1 , . . . , pn ) , q ¯ = (q1 , . . . , qn ) are nonnegative. Then one has the inequality n X i=1

2

pi |ai |

n X i=1

2

qi |bi | +

n X

2

pi |ci |

i=1

n X

qi |di |2

i=1

" ≥ 2 Re

n X

pi ai c¯i

i=1

n X

# qi bi d¯i . (1.16)

i=1

The case of equality for p ¯, q ¯ positive holds iff ai bj = ci dj for any i, j ∈ {1, . . . , n} . Proof. From the elementary inequality for complex numbers   |a|2 + |b|2 ≥ 2 Re a¯b , a, b ∈ C, with equality iff a = b, we have   ¯ |ai |2 |bj |2 + |ci |2 |dj |2 ≥ 2 Re ai c¯i bj dj

(1.17)

for any i, j ∈ {1, . . . , n} . Multiplying (1.17) by pi qj ≥ 0 and summing over i and j from 1 to n, we deduce (1.16). The case of equality is obvious and we omit the details. Remark 13 Similar particular cases may be stated but we omit the details.

1.4. A RELATED ADDITIVE INEQUALITY

1.4

7

A Related Additive Inequality

The following inequality was obtained in [4, Theorem 1.1]. ¯ = (b1 , . . . , bn ) are sequences of real Theorem 14 If ¯ a = (a1 , . . . , an ) , b ¯ = (d1 , . . . , dn ) are nonnegative, then numbers and ¯ c = (c1 , . . . , cn ) , d n X i=1

di

n X i=1

ci a2i

+

n X i=1

ci

n X

di b2i

≥2

i=1

n X

c i ai

n X

i=1

di bi .

(1.18)

i=1

If ci and di (i = 1, . . . , n) are positive, then equality holds in (1.18) iff ¯ a= ¯ ¯ ¯ b = k where k = (k, k, . . . , k) is a constant sequence. Proof. We will follow the proof from [4]. From the elementary inequality a2 + b2 ≥ 2ab for any a, b ∈ R

(1.19)

with equality iff a = b; we have a2i + b2j ≥ 2ai bj for any i, j ∈ {1, . . . , n} .

(1.20)

Multiplying (1.20) by ci dj ≥ 0, i, j ∈ {1, . . . , n} and summing over i from 1 to n and over j from 1 to n, we deduce (1.18). If ci , dj > 0 (i = 1, . . . , n) , then the equality holds in (1.18) iff ai = bj for any i, j ∈ {1, . . . , n} which is equivalent with the fact that ai = bi = k for any i ∈ {1, . . . , n} . The following corollary holds [4, p. 4]. ¯ are nonnegative sequences, then Corollary 15 If ¯ a and b " n # n n n n n X X X X 1 X 3X ai bi + ai b3i ≥ a2i b2i ; 2 i=1 i=1 i=1 i=1 i=1 i=1 " n # n n n X X 1 X X 2 ai a bi + bi b2i ai ≥ 2 i=1 i=1 i i=1 i=1

n X

(1.21)

!2 ai b i

i=1

Another corollary that may be obtained is [4, p. 4 – 5].

.

(1.22)

8

CHAPTER 1. (CBS) – TYPE INEQUALITIES

¯ are sequences of positive real numbers, then Corollary 16 If ¯ a and b Pn 1 Pn 1 n X a2i + b2i i=1 ai i=1 Pn 1 bi , ≥ (1.23) 2a b i i i=1 a b i=1 i i n X i=1

n n n X 1 X 1 X ai + bi ≥ 2n2 , b a i=1 i i=1 i i=1

and n

n X a2 + b 2 i

i=1

i

2a2i b2i

≥

(1.24)

n n X 1 X1 . a b i i i=1 i=1

(1.25)

The following version for complex numbers also holds. ¯ = (b1 , . . . , bn ) are sequences of complex Theorem 17 If ¯ a = (a1 , . . . , an ) , b numbers, then for p ¯ = (p1 , . . . , pn ) and q ¯ = (q1 , . . . , qn ) two sequences of nonnegative real numbers, one has the inequality " n # n n n n n X X X X X X 2 2 qi pi |ai | + pi qi |bi | ≥ 2 Re p i ai qi¯bi . (1.26) i=1

i=1

i=1

i=1

i=1

i=1

¯ = k ¯ = For p ¯, q ¯ positive sequences, the equality holds in (1.26) iff ¯ a = b (k, . . . , k) . The proof goes in a similar way with the one in Theorem 14 on making use of the following elementary inequality holding for complex numbers   |a|2 + |b|2 ≥ 2 Re a¯b , a, b ∈ C; (1.27) with equality iff a = b.

1.5

A Parameter Additive Inequality

The following inequality was obtained in [4, Theorem 4.1]. ¯ = (b1 , . . . , bn ) be sequences of real Theorem 18 Let ¯ a = (a1 , . . . , an ) , b ¯ numbers and ¯ c = (c1 , . . . , cn ) , d = (d1 , . . . , dn ) be nonnegative. If α, β > 0 and γ ∈ R such that γ 2 ≤ αβ, then α

n X i=1

di

n X i=1

a2i ci + β

n X i=1

ci

n X i=1

b2i di ≥ 2γ

n X i=1

ci ai

n X i=1

di bi .

(1.28)

1.5. A PARAMETER ADDITIVE INEQUALITY

9

Proof. We will follow the proof from [4]. Since α, β > 0 and γ 2 ≤ αβ, it follows that for any x, y ∈ R one has αx2 + βy 2 ≥ 2γxy.

(1.29)

Choosing in (1.29) x = ai , y = bj (i, j = 1, . . . , n) , we get αa2i + βb2j ≥ 2γai bj for any i, j ∈ {1, . . . , n} .

(1.30)

If we multiply (1.30) by ci dj ≥ 0 and sum over i and j from 1 to n, we deduce the desired inequality (1.28). The following corollary holds. ¯ are nonnegative sequences and α, β, γ are as in Corollary 19 If ¯ a and b Theorem 18, then α

n X i=1

α

n X i=1

ai

bi

n X

a3i + β

i=1 n X

a2i bi

+β

n X

n X

i=1

i=1

n X

n X

i=1

i=1

ai

bi

b3i ≥ 2γ

n X

a2i

n X

i=1

b2i ai

≥ 2γ

(1.31)

i=1 n X

i=1

b2i , !2

ai b i

.

(1.32)

i=1

The following particular case is important [4, p. 8]. ¯ be sequences of real numbers. If p Theorem 20 Let ¯ a, b ¯ is a sequence of Pn nonnegative real numbers with i=1 pi > 0, then: Pn Pn Pn n n X X 2 2 i=1 pi ai bi Pi=1 pi ai i=1 pi bi p i ai p i bi ≥ . (1.33) n i=1 pi i=1 i=1 In particular, n X i=1

a2i

n X i=1

n

b2i ≥

n

n

X X 1X ai b i ai bi . n i=1 i=1 i=1

(1.34)

Proof. We will follow the proof from [4, p. 8]. Pn 2 IfP we choose in Theorem 18, c = d = p (i = 1, . . . , n) and α = i i i i=1 pi bi , P n n β = i=1 pi a2i , γ = i=1 pi ai bi , we observe, by the (CBS) −inequality with the weights pi (i = 1, . . . , n) one has γ 2 ≤ αβ, and then by (1.28) we deduce (1.33).

10

CHAPTER 1. (CBS) – TYPE INEQUALITIES

¯ are asynchronous, i.e., Remark 21 If we assume that ¯ a and b (ai − aj ) (bi − bj ) ≤ 0 for any i, j ∈ {1, . . . , n} ˇ sev’s inequality then by Cebyˇ n X

p i ai

n X

i=1

p i bi ≥

i=1

n X

pi

n X

i=1

p i ai b i

(1.35)

i=1

respectively n X

ai

i=1

n X

bi ≥ n

i=1

n X

ai b i ,

(1.36)

i=1

we have the following refinements of the (CBS) −inequality Pn Pn Pn n n X X 2 2 i=1 pi ai bi Pi=1 pi ai i=1 pi bi p i ai pi bi ≥ n i=1 pi i=1 i=1 ! 2 n X ≥ p i ai b i

(1.37)

i=1

Pn

provided

i=1

pi ai bi ≥ 0, respectively

n X

a2i

i=1

n X i=1

n

n

n

X X 1X b2i ≥ ai b i ai bi ≥ n i=1 i=1 i=1

n X

!2 ai b i

(1.38)

i=1

provided

Pn

1.6

A Generalisation Provided by Young’s Inequality

i=1

ai bi ≥ 0.

The following result was obtained in [4, Theorem 5.1]. ¯ = (b1 , . . . , bn ) , p Theorem 22 Let ¯ a = (a1 , . . . , an ) , b ¯ = (p1 , . . . , pn ) and q ¯ = (q1 , . . . , qn ) be sequences of nonnegative real numbers and α, β > 1 with 1 + β1 = 1. Then one has the inequality α α

n X i=1

qi

n X i=1

pi bβi

+β

n X i=1

pi

n X i=1

qi aαi

≥ αβ

n X i=1

p i bi

n X i=1

qi ai .

(1.39)

1.6. A GENERALISATION PROVIDED BY YOUNG’S INEQUALITY 11 If p ¯ and q ¯ are sequences of positive real numbers, then the equality holds in (1.39) iff there exists a constant k ≥ 0 such that aαi = bβi = k for each i ∈ {1, . . . , n} . Proof. It is, by the Arithmetic-Geometric inequality [5, p. 15], well known that 1 1 1 1 1 1 x + y ≥ x α y β for x, y ≥ 0, + = 1, α, β > 1 α β α β

(1.40)

with equality iff x = y. Applying (1.40) for x = aαi , y = bβj (i, j = 1, . . . , n) we have αbβj + βaαi ≥ αβai bj for any i, j ∈ {1, . . . , n}

(1.41)

with equality iff aαi = bβj for any i, j ∈ {1, . . . , n} . If we multiply (1.41) by qi pj ≥ 0 (i, j ∈ {1, . . . , n}) and sum over i and j from 1 to n we deduce (1.39). The case of equality is obvious by the above considerations. The following corollary is a natural consequence of the above theorem. ¯ α and β be as in Theorem 22. Then Corollary 23 Let ¯ a, b, n n n n n n 1 X X α+1 1 X X β+1 X 2 X 2 bi a + ai b ≥ ai bi ; α i=1 i=1 i β i=1 i=1 i i=1 i=1 n n n n 1X X α 1X X β ai b i ai + bi ai b i ≥ α i=1 i=1 β i=1 i=1

n X

(1.42)

!2 ai b i

.

(1.43)

i=1

The following result which provides a generalisation of the (CBS) −inequality may be obtained by Theorem 22 as well [4, Theorem 5.2]. Theorem 24 Let x ¯ and y ¯ be sequences of positive real numbers. If α, β are as above, then ! n !2 n n n X X 1 X α 2−α 1 X β 2−β x y + x y · yi2 ≥ xi yi . (1.44) α i=1 i i β i=1 i i i=1 i=1 The equality holds iff x ¯ and y ¯ are proportional.

12

CHAPTER 1. (CBS) – TYPE INEQUALITIES

Proof. Follows by Theorem 22 on choosing pi = qi = yi2 , ai = i ∈ {1, . . . , n} .

xi , yi

bi =

xi , yi

Remark 25 For α = β = 2, we recapture the (CBS) −inequality. Remark 26 For ai = |zi | , bi = |wi | , with zi , wi ∈ C; i = 1, . . . , n, we may obtain similar inequalities for complex numbers. We omit the details.

1.7

Further Generalisations via Young’s Inequality

The following inequality is known in the literature as Young’s inequality pxq + qy p ≥ pqxy,

x, y ≥ 0 and

1 1 + = 1, p > 1 p q

(1.45)

with equality iff xq = y p . The following result generalising the (CBS) −inequality was obtained in [6, Theorem 2.1] (see also [7, Theorem 1]). Theorem 27 Let x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) be sequences of complex numbers and p ¯ = (p1 , . . . , pn ) , q ¯ = (q1 , . . . , qn ) be two sequences of nonnegative real numbers. If p > 1, p1 + 1q = 1, then n n n n X X 1X 1X p p q pk |xk | qk |yk | + qk |xk | pk |yk |q p k=1 q k=1 k=1 k=1

≥

n X k=1

pk |xk yk |

n X

qk |xk yk | . (1.46)

k=1

Proof. We shall follow the proof in [6]. Choosing x = |xj | |yi | , y = |xi | |yj | , i, j ∈ {1, . . . , n} , we get from (1.45) q |xi |p |yj |p + p |xj |q |yi |q ≥ pq |xi yi | |xj yj |

(1.47)

for any i, j ∈ {1, . . . , n} . Multiplying with pi qj ≥ 0 and summing over i and j from 1 to n, we deduce the desired result (1.46). The following corollary is a natural consequence of the above theorem [6, Corollary 2.2] (see also [7, p. 105]).

1.7. FURTHER GENERALISATIONS VIA YOUNG’S INEQUALITY 13 Corollary 28 If x ¯ and y ¯ are as in Theorem 27 and m ¯ = (m1 , . . . , mn ) is a sequence of nonnegative real numbers, then n n n n X X 1X 1X p p q mk |xk | mk |yk | + mk |xk | mk |yk |q p k=1 q k=1 k=1 k=1

≥

n X

!2 mk |xk yk |

, (1.48)

k=1

where p > 1,

1 p

+

1 q

= 1.

Remark 29 If in (1.48) we assume that mk = 1, k ∈ {1, . . . , n} , then we obtain [6, p. 7] (see also [7, p. 105]) !2 n n n n n X X X X 1X 1 |xk |p |yk |p + |xk |q |yk |q ≥ |xk yk | , (1.49) p k=1 q k=1 k=1 k=1 k=1 which, in the particular case p = q = 2 will provide the (CBS) −inequality. The second generalisation of the (CBS) −inequality via Young’s inequality is incorporated in the following theorem [6, Theorem 2.4] (see also [7, Theorem 2]). Theorem 30 Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 27. Then one has the inequality n n n n X X 1X 1X pk |xk |p qk |yk |q + qk |xk |q pk |yk |p p k=1 q k=1 k=1 k=1

≥

n X

p−1

pk |xk | |yk |

k=1

n X

qk |xk | |yk |q−1 . (1.50)

k=1

Proof. We shall follow the proof in [6]. |x | i| , we get Choosing in (1.45), x = |yjj | , y = |x |yi |  p

|xj | |yj |

q

 +q

for any yi 6= 0, i, j ∈ {1, . . . , n} .

|xi | |yi |

p ≥ pq

|xi | |xj | |yi | |yj |

(1.51)

14

CHAPTER 1. (CBS) – TYPE INEQUALITIES It is easy to see that (1.51) is equivalent to q |xi |p |yj |q + p |yi |p |xj |q ≥ pq |xi | |yi |p−1 |xj | |yj |q−1

(1.52)

for any i, j ∈ {1, . . . , n} . Multiplying (1.52) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequlality (1.50). The following corollary holds [6, Corollary 2.5] (see also [7, p. 106]). Corollary 31 Let x ¯, y ¯, m ¯ and p ¯, q ¯ be as in Corollary 28. Then n n n n X X 1X 1X p q q mk |xk | mk |yk | + mk |xk | mk |yk |p p k=1 q k=1 k=1 k=1

≥

n X

n X

p−1

mk |xk | |yk |

k=1

mk |xk | |yk |q−1 . (1.53)

k=1

Remark 32 If in (1.53) we assume that mk = 1, k ∈ {1, . . . , n} , then we obtain [6, p. 8] (see also [7, p. 106]) n n n n X X 1X 1X p q q |xk | |yk | + |xk | |yk |p p k=1 q k=1 k=1 k=1

≥

n X

p−1

|xk | |yk |

k=1

n X

|xk | |yk |q−1 , (1.54)

k=1

which, in the particular case p = q = 2 will provide the (CBS) −inequality. The third result is embodied in the following theorem [6, Theorem 2.7] (see also [7, Theorem 3]). Theorem 33 Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 27. Then one has the inequality n n n n X X 1X 1X pk |xk |p qk |yk |q + qk |xk |p pk |yk |q p k=1 q k=1 k=1 k=1

≥

n X k=1

pk |xk yk |

n X k=1

pk |xk |p−1 |yk |q−1 . (1.55)

1.7. FURTHER GENERALISATIONS VIA YOUNG’S INEQUALITY 15 Proof. We shall follow the proof in [6]. |yi | |xi | If we choose x = |y and y = |x in (1.45) we get j| j| q  p  |yi | |xi | |xi | |yi | p +q ≥ pq , |yj | |xj | |xj | |yj | for any xi , yj 6= 0, i, j ∈ {1, . . . , n} , giving q |xi |p |yj |q + p |yi |q |xj |p ≥ pq |xi yi | |xj |p−1 |yj |q−1

(1.56)

for any i, j ∈ {1, . . . , n} . Multiplying (1.56) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequality (1.55). The following corollary is a natural consequence of the above theorem [7, p. 106]. Corollary 34 Let x ¯, y ¯, m ¯ and p ¯, q ¯ be as in Corollary 28. Then one has the inequality: n X k=1

p

mk |xk |

n X

q

mk |yk | ≥

k=1

n X

mk |xk yk |

k=1

n X

mk |xk |p−1 |yk |q−1 .

(1.57)

k=1

Remark 35 If in (1.57) we assume that mk = 1, k = {1, . . . , n} , then we obtain [6, p. 8] (see also [7, p. 10]) n X k=1

p

|xk |

n X k=1

q

|yk | ≥

n X

|xk yk |

k=1

n X

|xk |p−1 |yk |q−1 ,

(1.58)

k=1

which, in the particular case p = q = 2 will provide the (CBS) −inequality. The fourth generalisation of the (CBS) −inequality is embodied in the following theorem [6, Theorem 2.9] (see also [7, Theorem 4]). Theorem 36 Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 27. Then one has the inequality n n n n X X 1X 1X pk |xk |2 qk |yk |q + pk |yk |2 qk |xk |p q k=1 p k=1 k=1 k=1

≥

n X k=1

qk |xk yk |

n X k=1

2

2

pk |xk | q |yk | p . (1.59)

16

CHAPTER 1. (CBS) – TYPE INEQUALITIES Proof. We shall follow the proof in [6]. 2 2 Choosing in (1.45), x = |xi | q |yj | , y = |xj | |yi | p , we get 2

2

p |xi |2 |yj |q + q |xj |p |yi |2 ≥ pq |xi | q |yi | p |xj yj |

(1.60)

for any i, j ∈ {1, . . . , n} . Multiply (1.60) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequality (1.60). The following corollary holds [6, Corollary 2.10] (see also [7, p. 107]). Corollary 37 Let x ¯, y ¯, m ¯ and p, q be as in Corollary 28. Then one has the inequality: n n n n X X 1X 1X 2 q 2 mk |xk | mk |yk | + mk |yk | mk |xk |p q k=1 p k=1 k=1 k=1

≥

n X

mk |xk yk |

n X

2

2

mk |xk | q |yk | p . (1.61)

k=1

k=1

Remark 38 If in (1.61) we take mk = 1, k ∈ {1, . . . , n} , then we get n n n n n n X X X X 2 2 1X 1X |xk |2 |yk |q + |yk |2 |xk |p ≥ |xk yk | |xk | q |yk | p , (1.62) q k=1 p k=1 k=1 k=1 k=1 k=1

which, in the particular case p = q = 2 will provide the (CBS) −inequality. The fifth result generalising the (CBS) −inequality is embodied in the following theorem [6, Theorem 2.12] (see also [7, Theorem 5]). Theorem 39 Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 27. Then one has the inequality n n n n X X 1X 1X pk |xk |2 qk |yk |q + pk |yk |2 qk |xk |p p k=1 q k=1 k=1 k=1

≥

n X k=1

2 p

pk |xk | |yk |

2 q

n X k=1

qk |xk |p−1 |yk |q−1 . (1.63)

1.7. FURTHER GENERALISATIONS VIA YOUNG’S INEQUALITY 17 Proof. We will follow the proof in [6]. 2

2

|xi | p |xj |

|yi | q |yj |

, yi , xj 6= 0, i, j ∈ {1, . . . , n} , we Choosing in (1.45), x = ,y= may write !p !q 2 2 2 2 |yi | q |xi | p |xi | p |yi | q ≥ pq , p +q |xj | |yj | |xj | |yj | from where results 2

2

p |yi |2 |xj |p + q |xi |2 |yj |q ≥ pq |xi | p |yi | q |xj |p−1 |yj |q−1

(1.64)

for any i, j ∈ {1, . . . , n} . Multiplying (1.64) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequality (1.63). The following corollary holds [6, Corollary 2.13] (see also [7, p. 108]). Corollary 40 Let x ¯, y ¯, m ¯ and p, q be as in Corollary 28. Then one has the inequality: n n n n X X 1X 1X 2 q 2 mk |xk | mk |yk | + mk |yk | mk |xk |p p k=1 q k=1 k=1 k=1

≥

n X

2 p

mk |xk | |yk |

k=1

2 q

n X

mk |xk |p−1 |yk |q−1 . (1.65)

k=1

Remark 41 If in (1.46) we choose mk = 1, k ∈ {1, . . . , n} , then we get [6, p. 10] (see also [7, p. 108]) n n n n X X 1X 1X |xk |2 |yk |q + |yk |2 |xk |p p k=1 q k=1 k=1 k=1

≥

n X k=1

2 p

|xk | |yk |

2 q

n X

|xk |p−1 |yk |q−1 , (1.66)

k=1

which in the particular case p = q = 2 will provide the (CBS) −inequality. Finally, the following result generalising the (CBS) −inequality holds [6, Theorem 2.15] (see also [7, Theorem 6]).

18

CHAPTER 1. (CBS) – TYPE INEQUALITIES

Theorem 42 Let x ¯, y ¯, p ¯, q ¯ and p, q be as in Theorem 27. Then one has the inequality: n n n n X X 1X 1X pk |xk |2 qk |yk |p + qk |yk |2 pk |xk |q p k=1 q k=1 k=1 k=1

≥

n X

2 p

pk |xk | |yk |

n X

2

qk |xk | q |yk | . (1.67)

k=1

k=1

Proof. We shall follow the proof in [6]. From (1.45) one has the inequality  p  q 2 2 2 2 p q q |xi | |yj | + p |xj | |yi | ≥ pq |xi | p |yi | |xj | q |yj |

(1.68)

for any i, j ∈ {1, . . . , n} . Multiplying (1.68) by pi qj ≥ 0 (i, j ∈ {1, . . . , n}) and summing over i and j from 1 to n, we deduce the desired inequality (1.67). The following corollary also holds [6, Corollary 2.16] (see also [7, p. 108]). Corollary 43 With the assumptions in Corollary 28, one has the inequality   n n X X 1 1 2 p q mk |xk | mk |yk | + |yk | p q k=1 k=1 ≥

n X

2

mk |xk | p |yk |

k=1

n X

2

mk |xk | q |yk | . (1.69)

k=1

Remark 44 If in (1.69) we choose mk = 1 (k ∈ {1, . . . , n}) , then we get  X n n  n n X X X 2 2 1 1 2 p q p |xk | |yk | + |yk | ≥ |xk | |yk | |xk | q |yk | , (1.70) p q k=1 k=1 k=1 k=1 which, in the particular case p = q = 2, provides the (CBS) −inequality.

1.8

A Generalisation Involving J−Convex Functions

For a > 1, we denote by expa the function expa : R → (0, ∞) , expa (x) = ax .

(1.71)

1.8. A GENERALISATION INVOLVING J−CONVEX FUNCTIONS 19 Definition 45 A function f : I ⊆ R → R is said to be J−convex on an interval I if   x+y f (x) + f (y) f ≤ for any x, y ∈ I. (1.72) 2 2 It is obvious that any convex function on I is a J convex function on I, but the converse does not generally hold. The following lemma holds (see [6, Lemma 4.3]). Lemma 46 Let f : I ⊆ R → R be a J−convex function on I, a > 1 and x, y ∈ R\ {0} with loga x2 , loga y 2 ∈ I. Then loga |xy| ∈ I and 
 
 {expb [f (loga |xy|)]}2 ≤ expb f loga x2 expb f loga y 2

(1.73)

for any b > 1. Proof. I, being an interval, is a convex set in R and thus loga |xy| =

 1 loga x2 + loga y 2 ∈ I. 2

Since f is J−convex, one has 


1 2 2 f (loga |xy|) = f loga x + loga y 2 f (loga x2 ) + f (loga y 2 ) ≤ . 2

(1.74)

Taking the expb in both parts, we deduce  f (loga x2 ) + f (loga y 2 ) expb [f (loga |xy|)] ≤ expb 2  
 
 12 = expb f loga x2 expb f loga y 2 , 

which is equivalent to (1.73). The following generalisation of the (CBS) −inequality in terms of a J− convex function holds [6, Theorem 4.4].

20

CHAPTER 1. (CBS) – TYPE INEQUALITIES

Theorem 47 Let f : I ⊆ R → R be a J−convex function on I, a, b > 1 ¯ = (b1 , . . . , bn ) sequences of nonzero real numbers. If and ¯ a = (a1 , . . . , an ) , b loga a2k , loga b2k ∈ I for all k ∈ {1, . . . , n} , then one has the inequality: ( n X

)2 expb [f (loga |ak bk |)]

k=1

≤

n X

n 
 X 
 2 expb f loga ak expb f loga b2k . (1.75)

k=1

k=1

Proof. Using Lemma 46 and the (CBS) −inequality one has n X

expb [f (loga |ak bk |)]

k=1 n X  
 
 1 ≤ expb f loga a2k expb f loga b2k 2 k=1

≤

n n n n X  
 1 o2  
 12 o2 X 2 expb f loga ak expb f loga b2k 2 k=1

! 12

k=1

which is clearly equivalent to (1.75). Remark 48 If in (1.75) we choose a = b > 1 and f (x) = x, x ∈ R, then we reapture the (CBS) −inequality.

1.9

A Functional Generalisation

The following result was proved in [9, Theorem 2]. Theorem 49 Let A be a subset of real numbers R, f : A → R and ¯ a = ¯ (a1 , . . . , an ) , b = (b1 , . . . , bn ) sequences of real numbers with the properties that (i) ai bi , a2i , b2i ∈ A for any i ∈ {1, . . . , n} , (ii) f (a2i ) , f (b2i ) ≥ 0 for any i ∈ {1, . . . , n} , (iii) f 2 (ai bi ) ≤ f (a2i ) f (b2i ) for any i ∈ {1, . . . , n} .

1.9. A FUNCTIONAL GENERALISATION

21

Then one has the inequality: "

n X

#2 f (ai bi )

≤

i=1

n X

f

a2i

n
X

i=1


f b2i .

(1.76)

i=1

Proof. We give here a simpler proof than that found in [9]. We have n X f (ai bi ) i=1

≤

n X

|f (ai bi )| ≤

i=1

n X 
 1 
 1 f a2i 2 f b2i 2 i=1

"

n  n  X 
 1 2 X 
 1 2 ≤ f a2i 2 f b2i 2 i=1

" =

n X i=1

# 12 (by the (CBS)-inequality)

i=1 n
X 2

f ai

# 21
2

f bi

i=1

and the inequality (1.76) is proved. Remark 50 It is obvious that for A = R and f (x) = x, we recapture the (CBS) −inequality. Assume that ϕ : N → N is Euler’s indicator. In 1940, T. Popoviciu [10] proved the following inequality for ϕ

[ϕ (ab)]2 ≤ ϕ a2 ϕ b2 for any natural number a, b (1.77) with equality iff a and b have the same prime factors. A simple proof of this fact may be done by using the representation     1 1 ϕ (n) = n 1 − ··· 1 − , p1 pk where n = pα1 1 pα2 2 · · · pαk k [8, p. 109]. The following generalisation of Popoviciu’s result holds [9, Theorem 1].

22

CHAPTER 1. (CBS) – TYPE INEQUALITIES

Theorem 51 Let ai , bi ∈ N (i = 1, . . . , n) . Then one has the inequality #2 " n n n X X
X
ϕ (ai bi ) ≤ ϕ a2i ϕ b2i . (1.78) i=1

i=1

i=1

Proof. Follows by Theorem 49 on taking into account that, by (1.77), [ϕ (ai bi )]2 ≤ ϕ (a2i ) ϕ (b2i ) for any i ∈ {1, . . . , n} . Further, let us denote by s (n) the sum of all relatively prime numbers with n and less than n. Then the following result also holds [9, Theorem 1]. Theorem 52 Let ai , bi ∈ N (i = 1, . . . , n) . Then one has the inequality " n #2 n n X X
X
2 s (ai bi ) ≤ s ai s b2i . (1.79) i=1

i=1

i=1

Proof. It is known (see for example [8, p. 109]) that for any n ∈ N one has

1 s (n) = nϕ (n) . 2

(1.80)

Thus



1 1 [s (ai bi )]2 = a2i b2i ϕ2 (ai bi ) ≤ a2i b2i ϕ a2i ϕ b2i = s a2i s b2i 4 4

(1.81)

for each i ∈ {1, . . . , n} . Using Theorem 49 we then deduce the desired inequality (1.79). The following corollaries of Theorem 49 are also natural to be considered [9, p. 126]. Corollary 53 Let ai , bi ∈ R (i = 1, . . . , n) and a > 1. Denote expa x = ax , x ∈ R. Then one has the inequality " n #2 n n X X
X
2 expa (ai bi ) ≤ expa ai expa b2i . (1.82) i=1

i=1

i=1

Corollary 54 Let ai , bi ∈ (−1, 1) (i = 1, . . . , n) and m > 0. Then one has the inequality: " n #2 n n X X X 1 1 1 ≤ . (1.83) m 2 m 2 m (1 − a b ) (1 − a ) (1 − b ) i i i i i=1 i=1 i=1

1.10. A GENERALISATION FOR POWER SERIES

1.10

23

A Generalisation for Power Series

The following result holds [11, Remark 2]. P k Theorem 55 Let F : (−r, r) → R, F (x) = ∞ k=0 αk x with αk ≥ 0, k ∈ N. ¯ = (b1 , . . . , bn ) are sequences of real numbers such that If ¯ a = (a1 , . . . , an ) , b ai bi , a2i , b2i ∈ (−r, r) for any i ∈ {1, . . . , n} ,

(1.84)

then one has the inequality: n X

n
X 2

F ai

i=1

"
2

F bi ≥

n X

i=1

#2 F (ai bi )

.

(1.85)

i=1

Proof. Firstly, let us observe that if x, y ∈ R such that xy, x2 , y 2 ∈ (−r, r) , then one has the inequality

[F (xy)]2 ≤ F x2 F y 2 . (1.86) Indeed, by the (CBS) −inequality, we have "

n X

#2 k k

αk x y

≤

n X

αk x

2k

k=0

k=0

n X

αk y 2k , n ≥ 0.

(1.87)

k=0

Taking the limit as n → ∞ in (1.87), we deduce (1.86). Using the (CBS) −inequality and (1.86) we have n n n X X X 
 1
 1  F (ai bi ) ≤ |F (ai bi )| ≤ F a2i 2 F b2i 2 i=1 i=1 i=1 ( n )1  n   2 X  1 2 X 1 2

 F b2i 2 ≤ F a2i 2 i=1

" =

n X i=1

i=1


2

F ai

n X

# 12
2

F bi

,

i=1

which is clearly equivalent to (1.85). The following particular inequalities of (CBS) −type hold [11, p. 164].

24

CHAPTER 1. (CBS) – TYPE INEQUALITIES ¯ are sequences of real numbers, then one has the inequality 1. If ¯ a, b " n #2 n n X X X

exp a2k exp b2k ≥ exp (ak bk ) ; (1.88) k=1 n X

k=1


2

sinh ak

n X

k=1

k=1

n X

n
X 2

cosh ak

k=1

k=1

"
2

sinh bk ≥

n X

#2 sinh (ak bk )

;

(1.89)

.

(1.90)

k=1

"
2

cosh bk ≥

k=1

n X

#2 cosh (ak bk )

k=1

¯ are such that ai , bi ∈ (−1, 1) , i ∈ {1, . . . , n} , then one has the 2. If ¯ a, b inequalities " n #2 n n X X X

tan a2k tan b2k ≥ tan (ak bk ) ; (1.91) k=1 n X

k=1


2

arcsin ak

k=1

"

n X

 n  Y 1 + a2 k=1


2

arcsin bk ≥

k=1

#

k

ln

k=1

"

1 − a2k

"

n X

#2 arcsin (ak bk )

 n  Y 1 + b2 k

ln

k=1

1 − b2k ≥

ln

 n  Y 1 + ak b k k=1

ln

n  Y k=1

1 1 − a2k

#

" ln

n  Y k=1

1 1 − b2k ≥

ln

n  Y k=1

k=1

#)2

1 − ak b k

; (1.93)

#

( "

n X

(1.92)

#

( "

"

;

k=1

1 1 − ak b k

#)2 ; (1.94)

" n #2 n X X 1 1 1 , m > 0. (1.95) m m ≥ m (1 − a (1 − a2k ) k=1 (1 − b2k ) k bk ) k=1

1.11. A GENERALISATION OF CALLEBAUT’S INEQUALITY

1.11

25

A Generalisation of Callebaut’s Inequality

The following result holds (see also [11, Theorem 2] for a generalisation for positive linear functionals). P k Theorem 56 Let F : (−r, r) → R, F (x) = ∞ k=0 αk x with αk ≥ 0, k ∈ ¯ = (b1 , . . . , bn ) are sequences of nonnegative real N. If ¯ a = (a1 , . . . , an ) , b numbers such that , a2−α bαi ∈ (0, r) for any i ∈ {1, . . . , n} ; α ∈ [0, 2] , ai bi , aαi b2−α i i

(1.96)

then one has the inequality "

n X

#2 ≤

F (ai bi )

i=1

n X

F aαi b2−α i

n
X

i=1


F ai2−α bαi .

(1.97)

i=1

Proof. Firstly, we note that for any x, y > 0 such that xy, xα y 2−α , x y ∈ (0, r) one has 2−α α



[F (xy)]2 ≤ F xα y 2−α F x2−α y α .

(1.98)

Indeed, using Callebaut’s inequality, i.e., we recall it [3] m X

!2 α i xi yi

i=1

≤

m X

αi xαi yi2−α

m X

αi x2−α yiα , i

(1.99)


i αi x2−α y α .

(1.100)

i=1

i=1

we may write, for m ≥ 0, that m X i=0

!2 i i

αi x y

≤

m X i=0


α 2−α i

αi x y

m X i=0

Taking the limit as m → ∞, we deduce (1.98).

26

CHAPTER 1. (CBS) – TYPE INEQUALITIES Using the (CBS) −inequality and (1.98) we may write: n X F (ai bi ) i=1

n X

≤

n X 
 21 
 12 2−α α |F (ai bi )| ≤ F aαi b2−α F a b i i i

i=1

i=1

( n )1  n   2 X  1 2 X 1 2

 2 ≤ F aαi b2−α F a2−α bαi 2 i i i=1

" =

n X

i=1 n
X 2−α

F aαi bi

i=1

# 21 F a2−α bi i


α

i=1

which is clearly equivalent to (1.97). The following particular inequalities also hold [11, pp. 165-166]. ¯ be sequences of nonnegative real numbers. Then one has 1. Let ¯ a and b the inequalities "

n X

#2 ≤

exp (ak bk )

k=1

"

n X

n X

exp aαk b2−α k

n
X

k=1

#2 ≤

sinh (ak bk )

k=1

"

n X

n X

≤

cosh (ak bk )

k=1

n X

(1.101)

k=1

sinh

aαk b2−α k

n
X

k=1

#2


exp a2−α bαk ; k
bαk ; sinh a2−α k

(1.102)


cosh ak2−α bαk .

(1.103)

k=1

cosh

aαk b2−α k

k=1

n
X k=1

¯ be such that ak , bk ∈ (0, 1) for any k ∈ {1, . . . , n} . Then 2. Let ¯ a and b one has the inequalities: "

n X k=1

#2 tan (ak bk )

≤

n X k=1

tan

aαk b2−α k

n
X k=1


α tan a2−α b ; k k

(1.104)

1.12. WAGNER’S INEQUALITY FOR REAL NUMBERS "

n X

27

#2 arcsin (ak bk )

k=1

≤

n X

arcsin

aαk b2−α k

ln

 n  Y 1 + ak b k k=1


arcsin ak2−α bαk ; (1.105)

k=1

k=1

( "

n
X

#)2

1 − ak b k " n  # " n  # Y 1 + a2−α bα  Y 1 + aα b2−α  k k k k ≤ ln ln ; (1.106) 1 − aαk b2−α 1 − a2−α bαk k k k=1 k=1

#)2 1 ln 1 − ak b k k=1 " n  # " Y # n  Y 1 1 ≤ ln ln . (1.107) 1 − aαk b2−α 1 − a2−α bαk k k k=1 k=1

( "

1.12

n  Y

Wagner’s Inequality for Real Numbers

The following generalisation of the (CBS) −inequality for sequences of real numbers is known in the literature as Wagner’s inequality [14], or [13] (see also [3, p. 85]). ¯ = (b1 , . . . , bn ) be sequences of real Theorem 57 Let ¯ a = (a1 , . . . , an ) and b numbers. If 0 ≤ x ≤ 1, then one has the inequality n X

!2 ak b k + x

k=1

X

ai b j

1≤i6=j≤n

" ≤

n X k=1

#" a2k + 2x

X 1≤i<j≤n

ai aj

n X k=1

# b2k + 2x

X

bi bj . (1.108)

1≤i<j≤n

Proof. We shall follow the proof in [12] (see also [3, p. 85]).

28

CHAPTER 1. (CBS) – TYPE INEQUALITIES For any x ∈ [0, 1] , consider the quadratic polynomial in y "

n X

P (y) := (1 − x)

2

(ak y − bk ) + x

n X

= (1 − x) y 2

n X

a2k − 2y

n X

k=1



ak b k + n X

− 2y

ak

n X

− 2y (1 − x)

n X

a2k + x

ak b k + x

=



n X

b2k + x n X

a2k + x 

− 2y

−

ak

+

k=1

ak b k + x 

b2k + x 

k=1

ak

n X

n X

n X

ak

k=1 !2

bk

  2 ak y2 

n X

bk −

k=1

−

bk

#

bk n X

n X

n X

!# ak b k

k=1

 b2k  .

k=1

k=1

Since, it is obvious that: n X

n X k=1

!2 ak

k=1 n X

ak

k=1

−

n X

a2k = 2

k=1

bk −

n X k=1

ak b k =

X

ai aj ,

1≤i<j≤n

X 1≤i6=j≤n

n X k=1

!2

k=1

k=1 n X

bk

k=1

k=1 !2

k=1

k=1

+

n X k=1

n X

ak

!

 y2

ak

n X



"

n X

!2 

k=1

k=1

 n X

!

k=1

n X

+ (1 − x)

b2k

k=1

k=1

"

#

k=1

!2

k=1

= (1 − x)

n X

k=1

n X

+ x y 2 

(ak y − bk )

k=1

k=1

"

#2

ai b j

!2  bk



1.13. WAGNER’S INEQUALITY FOR COMPLEX NUMBERS

29

and n X

!2 −

bk

n X

bi bj ,

1≤i<j≤n

k=1

k=1

X

b2k = 2

we get

P (y) =

n X

! a2k

+ 2x

X

y2

ai aj

1≤i<j≤n

k=1

− 2y

n X

! X

ak b k + x

k=1

ai b j

1≤i6=j≤n

+

n X

b2k + 2x

k=1

X

bi bj .

1≤i<j≤n

Taking into consideration, by the definition of P, that P (y) ≥ 0 for any y ∈ R, it follows that the discriminant ∆ ≤ 0, i.e., 1 0≥ ∆= 4

n X

!2 ak b k + x

k=1

X

ai b j

1≤i6=j≤n

−

n X k=1

! a2k + 2x

X 1≤i<j≤n

ai aj

n X k=1

! b2k + 2x

X

bi bj

1≤i<j≤n

and the inequality (1.108) is proved. Remark 58 If x = 0, then from (1.108) we recapture the (CBS) −inequality for real numbers.

1.13

Wagner’s inequality for Complex Numbers

The following inequality which provides a version for complex numbers of Wagner’s result holds [15].

30

CHAPTER 1. (CBS) – TYPE INEQUALITIES

¯ = (b1 , . . . , bn ) be sequences of comTheorem 59 Let ¯ a = (a1 , . . . , an ) and b plex numbers. Then for any x ∈ [0, 1] one has the inequality  2 n X

 X Re ak¯bk + x Re ai¯bj   k=1

1≤i,j≤n i6=j

" ≤

n X

# X

2

|ak | + 2x

Re (ai a ¯j )

1≤i<j≤n

k=1

" ×

n X

# X

|bk |2 + 2x

Re bi¯bj




. (1.109)

1≤i<j≤n

k=1

Proof. Start with the function f : R → R, n 2 n X X f (t) = (1 − x) |tak − bk |2 + x (tak − bk ) . k=1

(1.110)

k=1

We have f (t) = (1 − x)

n X

(tak − bk ) t¯ ak − ¯bk




(1.111)

k=1

+x t

n X

ak −

k=1

" = (1 − x) t2

n X

k=1 n X

! bk

t

+ x t2

n X

|ak | − t

k=1



! ¯bk

k=1

2

n X

k=1 n X

k=1 n X

n X

k=1

k=1

|ak | − t

2

a ¯k −

n X

k=1 n X

k=1

"

n X

k=1

bk

bk a ¯k − t a ¯k − t

k=1

n X

ak¯bk +

n X

ak

¯bk +

n X

2  n X 2  = (1 − x) |ak | + x ak  t2 k=1 k=1 " " n ## n n X X X
¯bk t + 2 (1 − x) Re ak¯bk + x Re ak k=1

+ (1 − x)

k=1

k=1

2 n X |bk |2 + x bk . k=1

|bk |

k=1

n X

n X

# 2

k=1

k=1

# 2

|bk |

1.13. WAGNER’S INEQUALITY FOR COMPLEX NUMBERS

31

Observe that 2 n n n X X X ak = ai a ¯j = |ai |2 + i,j=1

k=1

= =

n X

i=1

X

|ai |2 +

i=1

1≤i<j≤n

n X

X

|ai |2 + 2

i=1

X

ai a ¯j

(1.112)

1≤i6=j≤n

X

ai a ¯j +

ai a ¯j

1≤j
Re (ai a ¯j )

1≤i<j≤n

and, similarly, 2 n n X X X
bk = |bi |2 + 2 Re bi¯bj . i=1

k=1

(1.113)

1≤i<j≤n

Also n X

ak

n X

¯bk =

X

ai¯bi +

i=1

k=1

k=1

n X

ai¯bj

1≤i,j≤n i6=j

and thus Re

n X k=1

ak

n X

! ¯bk

=

n X

X

Re ai¯bi + Re ai¯bj .

i=1

k=1

(1.114)

1≤i,j≤n i6=j

Utilising (1.112) – (1.114), by (1.111), we deduce " f (t) =

n X

# X

2

|ak | + 2x

Re (ai a ¯j ) t2

1≤i<j≤n

k=1

  n X

 X Re ak¯bk + x Re ai¯bj  t +2  k=1

+

n X k=1

|bk |2 + 2x

1≤i,j≤n i6=j

X 1≤i<j≤n


Re bi¯bj .

(1.115)

32

CHAPTER 1. (CBS) – TYPE INEQUALITIES

Since, by (1.110), f (t) ≥ 0 for any t ∈ R, it follows that the discriminant of the quadratic function given by (1.115) is negative, i.e., 1 ∆ 4  2 n X

 X =  Re ak¯bk + x Re ai¯bj 

0 ≥

k=1

" −

n X

1≤i,j≤n i6=j

#" |ak |2 + 2x

X

Re (ai a ¯j )

1≤i<j≤n

k=1

n X

# |bk |2 + 2x

X

Re bi¯bj




1≤i<j≤n

k=1

and the inequality (1.109) is proved. Remark 60 If x = 0, then we get the (CBS) −inequality "

n X k=1

#2 Re ak¯bk




≤

n X k=1

|ak |2

n X k=1

|bk |2 .

(1.116)

Bibliography [1] V. BUNIAKOWSKI, Sur quelques in´egalit´es concernant les int´egrales aux differences finies, Mem. Acad. St. Petersburg, (7) 1 (1859), No. 9, 1-18. ´ [2] A.L. CAUCHY, Cours d’Analyse de l’Ecole Royale Polytechnique, I re Partie, Analyse Alg´ebrique, Paris, 1821. ´ J.E. PECARI ˇ ´ and A.M. FINK, Classical and [3] D.S. MITRINOVIC, C New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993. [4] S.S. DRAGOMIR, On some inequalities (Romanian), “Caiete Metodico S¸tiint¸ifice”, No. 13, 1984, pp. 20. Faculty of Mathematics, Timi¸soara University, Romania. [5] E.F. BECKENBACH and R. BELLMAN, Inequalities, Springer-Verlag, Berlin-G¨ottingen-Heidelberg, 1961. [6] S.S. DRAGOMIR, On Cauchy-Buniakowski-Schwartz’s Inequality for Real Numbers (Romanian), “Caiete Metodico-S¸tiint¸ifice”, No. 57, pp. 24, 1989. Faculty of Mathematics, Timi¸soara University, Romania. ´ [7] S.S. DRAGOMIR and J. SANDOR, Some generalisations of CauchyBuniakowski-Schwartz’s inequality (Romanian), Gaz. Mat. Metod. (Bucharest), 11 (1990), 104-109. [8] I. CUCUREZEANU, Problems on Number Theory (Romanian), Ed. Technicˇa, Bucharest, 1976. [9] S.S. DRAGOMIR, On an inequality of Tiberiu Popoviciu (Romanian), Gaz. Mat. Metod., (Bucharest) 8 (1987), 124-128. ZBL No. 712:110A. 33

34

BIBLIOGRAPHY

[10] T. POPOVICIU, Gazeta Matematicˇ a, 16 (1940), p. 334. [11] S.S. DRAGOMIR, Inequalities of Cauchy-Buniakowski-Schwartz’s type for positive linear functionals (Romanian), Gaz. Mat. Metod. (Bucharest), 9 (1988), 162-164. [12] T. ANDRESCU, D. ANDRICA and M.O. DRˆIMBE, The trinomial principle in obtaining inequalities (Romanian), Gaz. Mat. (Bucharest), 90 (1985), 332-338. ¨ [13] P. FLOR, Uber eine Unglichung von S.S. Wagner, Elemente Math., 20 (1965), 136. [14] S.S. WAGNER, Amer. Math. Soc., Notices, 12 (1965), 220. [15] S.S. DRAGOMIR, A version of Wagner’s inequality for complex numbers, submitted.

Chapter 2 Refinements of the (CBS) −Inequality 2.1

A Refinement in Terms of Moduli

The following result was proved in [1]. ¯ = (b1 , . . . , bn ) be sequences of real Theorem 61 Let ¯ a = (a1 , . . . , an ) and b numbers. Then one has the inequality n X k=1

a2k

n X k=1

b2k −

n X

!2 ak b k

k=1

n n n n X X X X ≥ ak |ak | bk |bk | − ak |bk | |ak | bk ≥ 0. (2.1) k=1

k=1

k=1

k=1

Proof. We will follow the proof from [1]. For any i, j ∈ {1, . . . , n} the next elementary inequality is true: |ai bj − aj bi | ≥ ||ai bj | − |aj bi || .

(2.2)

By multiplying this inequality with |ai bj − aj bi | ≥ 0 we get (ai bj − aj bi )2 ≥ |(ai bj − aj bi ) (|ai | |bj | − |aj | |bi |)| (2.3) = |ai |ai | bj |bj | + bi |bi | aj |aj | − |ai | bi aj |bj | − ai bj |aj | |bi || . 35

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

36

Summing (2.3) over i and j from 1 to n, we deduce n X

(ai bj − aj bi )2

i,j=1

n X ai |ai | bj |bj | + bi |bi | aj |aj | − |ai | bi aj |bj | − ai bj |aj | |bi | ≥ i,j=1 n X ≥ (ai |ai | bj |bj | + bi |bi | aj |aj | − |ai | bi aj |bj | − ai bj |aj | |bi |) , i,j=1

giving the desired inequality (2.1). The following corollary is a natural consequence of (2.1) [1, Corollary 4]. Corollary 62 Let ¯ a be a sequence of real numbers. Then n

n

1X 2 a − n k=1 k

1X ak n k=1

!2

n n n 1 X 1X 1X ≥ ak |ak | − ak · |ak | ≥ 0. (2.4) n n k=1 n k=1 k=1

There are some particular inequalities that may also be deduced from the above Theorem 61 (see [1, p. 80]). ¯ sequences of real numbers, one has sgn (ak ) = 1. Suppose that for ¯ a and b sgn (bk ) = ek ∈ {−1, 1} . Then one has the inequality n X

a2k

k=1

n X

b2k −

k=1

n X

!2 ak b k

k=1

X n X n ≥ ek a2k ek b2k − k=1 k=1

!2 ek ak bk ≥ 0. (2.5) k=1

n X

2. If ¯ a = (a1 , . . . , a2n ) , then we have the inequality 2n

2n X k=1

" a2k −

2n X k=1

#2 (−1)k ak

2n 2n X X ≥ ak (−1)k |ak | ≥ 0. k=1

k=1

(2.6)

2.1. A REFINEMENT IN TERMS OF MODULI

37

3. If ¯ a = (a1 , . . . , a2n+1 ) , then we have the inequality

(2n + 1)

2n+1 X

a2k −

k=1

2n+1 X

!2 (−1)k ak

k=1

2n+1 2n+1 X X ≥ ak (−1)k |ak | ≥ 0. (2.7) k=1

k=1

The following version for complex numbers is valid as well. ¯ = (b1 , . . . , bn ) be sequences of comTheorem 63 Let ¯ a = (a1 , . . . , an ) and b plex numbers. Then one has the inequality n X i=1

2

|ai |

n X i=1

2 n X |bi | − ai b i i=1 n n n n X X X X ≥ |ai | a ¯i |bi | bi − |ai | bi |bi | a ¯i ≥ 0. (2.8) 2

i=1

i=1

i=1

i=1

Proof. We have for any i, j ∈ {1, . . . , n} that |¯ ai b j − a ¯j bi | ≥ ||ai | |bj | − |aj | |bi || . Multiplying by |¯ ai b j − a ¯j bi | ≥ 0, we get |¯ ai b j − a ¯j bi |2 ≥ ||ai | a ¯i |bj | bj + |aj | a ¯j |bi | bi − |ai | bi |bj | a ¯j − |bi | a ¯i |aj | bj | . Summing over i and j from 1 to n and using the Lagrange’s identity for complex numbers: 2 n n n n X X X 1X 2 2 |ai | |bi | − |¯ ai b j − a ¯j bi |2 ai b i = 2 i=1

i=1

i=1

i,j=1

we deduce the desired inequality (2.8). Remark 64 Similar particular inequalities may be stated, but we omit the details.

38

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

2.2

A Refinement for a Sequence Whose Norm is One

The following result holds [1, Theorem 6]. ¯ = (b1 , . . . , bn ) be sequences of real Theorem 65 Let ¯ a = (a1 , . . . , an ), b P numbers and ¯ e = (e1 , . . . , en ) be such that ni=1 e2i = 1. Then the following inequality holds n X

a2i

i=1

n X

b2i

(2.9)

i=1

#2 " n n n n n X X X X X ≥ ak b k − ek ak ek bk + ek ak ek bk k=1 k=1 k=1 k=1 k=1 ! 2 n X ≥ ak b k . k=1

Proof. We will follow the proof from [1]. From the (CBS) −inequality, one has n X

"

n X

ak −

!

#2

ei ai ek

i=1

k=1

≥ Pn

2 k=1 ek n X

n X

bk −

ak −

n X

#2

! ei bi ek

i=1

! ei ai ek

#" bk −

i=1

n X

#)2

! ei bi ek

. (2.10)

i=1

= 1, a simple calculation shows that " ak −

n X

n X

!

#2

ei ai ek

=

i=1

k=1

k=1

"

k=1

( n " X k=1

Since

n X

" bk −

n X i=1

n X

a2k −

k=1

! ei bi ek

#2 =

n X k=1

n X

!2 ek ak

,

k=1

b2k

−

n X k=1

!2 ek bk

,

2.2. A REFINEMENT FOR A SEQUENCE WHOSE NORM IS ONE 39 and n X

" ak −

n X

#"

!

n X

bk −

ei ai ek

i=1

k=1

#

! ei bi ek

i=1

=

n X

ak b k −

k=1

n X

ek ak

k=1

n X

ek bk

k=1

and then the inequality (2.10) becomes  !2   n !2  n n n X X X X  a2k − ek ak   b2k − ek bk  k=1

k=1

k=1

k=1 n X

≥

ak b k −

k=1

n X

ek ak

k=1

Using the elementary inequality

m2 − l2 p2 − q 2 ≤ (mp − lq)2 ,

n X

!2 ≥ 0. (2.11)

ek bk

k=1

m, l, p, q ∈ R

for the choices n X

m=

! 21 a2k

,

n X l= ek ak ,

p=

k=1

nk=1 X and q = ek bk

n X

! 12 b2k

k=1

k=1

the above inequality (2.11) provides the following result 

n X

 k=1

! 21 a2k

n X

! 12 b2k

k=1

2 n n X X − ek ak ek bk  k=1 k=1 2 n n n X X X ≥ ak b k − ek ak ek bk . (2.12) k=1

Since n X k=1

! 21 a2k

n X k=1

! 12 b2k

k=1

k=1

n n X X ≥ ek ak ek bk k=1

k=1

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

40

then, by taking the square root in (2.12) we deduce the first part of (2.9). The second part is obvious, and the theorem is proved. The following corollary is a natural consequence of the above theorem [1, Corollary 7]. ¯ ¯ Corollary 66 Let ¯ a, b, e be as in Theorem 65. If has the inequality: n X

n X

a2k

k=1

b2k ≥ 4

n X

k=1

!2 ek ak

k=1

Pn

n X

k=1

ak bk = 0, then one

!2 ek bk

.

(2.13)

k=1

The following inequalities are interesting as well [1, p. 81]. ¯ one has the inequality 1. For any ¯ a, b n X k=1

a2k

n X

b2k

(2.14)

k=1

#2 " n n n n n X X X X X 1 1 ≥ ak b k − ak bk + ak bk n k=1 k=1 n k=1 k=1 k=1 ! 2 n X ≥ ak b k . k=1

2. If

Pn

k=1

ak bk = 0, then n X k=1

a2k

n X k=1

b2k ≥

4 n2

n X k=1

!2 ak

n X

!2 bk

.

(2.15)

k=1

In a similar manner, we may state and prove the following result for complex numbers. ¯ = (b1 , . . . , bn ) be sequences of complex Theorem 67 Let ¯ a = (a1 , . . . , an ), b numbers and ¯ e = (e1 , . . . , en ) a sequence of complex numbers satisfying the

2.3. A SECOND REFINEMENT IN TERMS OF MODULI condition holds

Pn

2 i=1 ei

n X

2

|ai |

i=1

41

= 1. Then the following refinement of the (CBS) −inequality n X

|bi |2

(2.16)

i=1

#2 " n n n n n X X X X X ¯ ¯ ¯ ≥ ak b k − ak e¯k · ek bk + ak e¯k · ek bk k=1 k=1 k=1 k=1 k=1 n 2 X ak¯bk . ≥ k=1

The proof is similar to the one in Theorem 65 on using the corresponding (CBS) −inequality for complex numbers. Remark 68 Similar particular inequalities may be stated, but we omit the details.

2.3

A Second Refinement in Terms of Moduli

The following lemma holds. Lemma 69 Let ¯ a = (a1 , . . . , an ) be a sequence of real ¯ = Pn numbers and p (p1 , . . . , pn ) a sequence of positive real numbers with i=1 pi = 1. Then one has the inequality: !2 n n n n n X X X X X 2 p i ai − p i ai ≥ pi |ai | ai − pi |ai | p i ai . (2.17) i=1

i=1

i=1

i=1

i=1

Proof. By the properties of moduli we have (ai − aj )2 = |(ai − aj ) (ai − aj )| ≥ |(|ai | − |aj |) (ai − aj )| for any i, j ∈ {1, . . . , n} . This is equivalent to a2i − 2ai aj + a2j ≥ ||ai | ai + |aj | aj − |ai | aj − |aj | ai | for any i, j ∈ {1, . . . , n} .

(2.18)

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

42

If we multiply (2.18) by pi pj ≥ 0 and sum over i and j from 1 to n we deduce n X j=1

pj

n X

pi a2i

−2

n X i=1

i=1 n X

≥

p i ai

n X

p j aj +

j=1

n X

pi

i=1

n X

pj a2j

j=1

pi pj |ai | ai + |aj | aj − |ai | aj − |aj | ai

i,j=1

n X ≥ pi pj (|ai | ai + |aj | aj − |ai | aj − |aj | ai ) , i,j=1

which is clearly equivalent to (2.17). Using the above lemma, we may prove the following refinement of the (CBS) -inequality. ¯ = (b1 , . . . , bn ) be two sequences of Theorem 70 Let ¯ a = (a1 , . . . , an ) and b real numbers. Then one has the inequality n X

a2i

i=1

n X i=1

b2i −

n X

!2 ai bi

i=1 n n n n X X X X ≥ a2i · sgn (ai ) |bi | bi − |ai bi | ai bi ≥ 0. (2.19) i=1

i=1

i=1

i=1

Proof. If we choose (for ai 6= 0, i ∈ {1, . . . , n}) in (2.17), that a2 pi := Pn i

2 k=1 ak

, xi =

bi , i ∈ {1, . . . , n} , ai

we get n X i=1

!2  2 n X bi a2i bi Pn Pn · − · 2 2 ai a ai k k=1 ak k=1 i=1 n n X bi bi X a2i a2i · − Pn P ≥ n 2 2 ai k=1 ak ai k=1 ak a2i

i=1

i=1

n 2 bi X a b i i P · n 2 ai a a i k k=1 i=1

2.3. A SECOND REFINEMENT IN TERMS OF MODULI

43

from where we get n X i=1

Pn |a | P Pn Pn i 2 ( ni=1 ai bi ) |a b | a b i=1 ai |bi | bi i i i i i=1 i=1 Pn Pn − Pn − Pn 2 ≥ 2 2 2 2 2 ( k=1 ak ) ( k=1 ak ) k=1 ak k=1 ak b2i

which is clearly equivalent to (2.19). The case for complex numbers is as follows. Lemma 71 Let ¯ z = (z1 , . . . , zn ) be a sequence of complex ¯= Pn numbers and p (p1 , . . . , pn ) a sequence of positive real numbers with i=1 pi = 1. Then one has the inequality: n X i=1

n 2 n n n X X X X pi |zi |2 − pi zi ≥ pi |zi | zi − pi |zi | pi zi . i=1

i=1

i=1

(2.20)

i=1

Proof. By the properties of moduli for complex numbers we have |zi − zj |2 ≥ |(|zi | − |zj |) (zi − zj )| for any i, j ∈ {1, . . . , n} , which is clearly equivalent to |zi |2 − 2 Re (zi z¯j ) + |zj |2 ≥ ||zi | zi + |zj | zj − zi |zj | − |zi | zj | for any i, j ∈ {1, . . . , n} . If we multiply with pi pj ≥ 0 and sum over i and j from 1 to n, we deduce the desired inequality (2.20). Now, in a similar manner to the one in Theorem 70, we may state the following result for complex numbers. ¯ = (b1 , . . . , bn ) Theorem 72 Let ¯ a = (a1 , . . . , an ) (ai = 6 0, i = 1, . . . , n)and b be two sequences of complex numbers. Then one has the inequality: n X i=1

2 n n X X |ai |2 |bi |2 − a ¯ i bi i=1 i=1 n n n X |a | X X i |bi | bi − |ai | bi a ¯i bi ≥ 0. (2.21) ≥ a i i=1 i=1 i=1

44

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

2.4

A Refinement for a Sequence Less than the Weights

The following result was obtained in [1, Theorem 9] (see also [2, Theorem 3.10]). ¯ = (b1 , . . . , bn ) be sequences of real Theorem 73 Let ¯ a = (a1 , . . . , an ), b numbers and p ¯ = (p1 , . . . , pn ), q ¯ = (q1 , . . . , qn ) be sequences of nonnegative real numbers such that pk ≥ qk for any k ∈ {1, . . . , n} . Then we have the inequality n X

pk a2k

k=1

n X

pk b2k

(2.22)

k=1

 ! 21 2 n n n X X X (pk − qk ) ak bk + qk a2k qk b2k  ≥  k=1 k=1 k=1 #2 " n n X X ≥ (pk − qk ) ak bk + qk ak bk k=1 k=1 ! 2 n X ≥ p k ak b k . k=1

Proof. We shall follow the proof in [1]. Since pk −qk ≥ 0, then the (CBS) −inequality for the weights rk := pk −qk will give n X k=1

pk a2k

−

n X k=1

! qk a2k

n X

pk b2k

−

k=1

n X

! qk b2k

k=1

" ≥

n X

#2 (pk − qk ) ak bk

k=1

Using the elementary inequality (ac − bd)2 ≥ a2 − b2





c2 − d2 ,

a, b, c, d ∈ R

. (2.23)

2.4. A REFINEMENT FOR A SEQUENCE LESS THAN THE WEIGHTS45 for the choices n X

a=

! 21 pk a2k

n X

, b=

k=1

! 12 qk a2k

, c=

k=1

d=

! 12 pk b2k

k=1

and n X

n X

! 12 qk b2k

k=1

we deduce by (2.23) that 

n X



! 21 pk a2k

k=1

! 12

n X

pk b2k

−

n X

k=1

! 12

n X

qk a2k

k=1

! 12 2 qk b2k



k=1

" ≥

n X

#2 (pk − qk ) ak bk

. (2.24)

k=1

Since, obviously, n X

! 12 pk a2k

k=1

n X

! 12 pk b2k

≥

k=1

n X

! 12 qk a2k

k=1

n X

! 12 qk b2k

k=1

then, by (2.24), on taking the square root, we would get n X k=1

! 12 pk a2k

n X

! 12 pk b2k

k=1

≥

n X k=1

! 12 qk a2k

n X k=1

! 12 qk b2k

n X + (pk − qk ) ak bk , k=1

which provides the first inequality in (2.22). The other inequalities are obvious and we omit the details. The following corollary is a natural consequence of the above theorem [2, Corollary 3.11].

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

46

¯ be sequences of real numbers and ¯ Corollary 74 Let ¯ a, b s = (s1 , . . . , sn ) be such that 0 ≤ sk ≤ 1 for any k ∈ {1, . . . , n} . Then one has the inequalities  ! 21 2 n n n X X X sk b2k  (1 − sk ) ak bk + sk a2k a2k b2k ≥  k=1 k=1 k=1 k=1 k=1 #2 " n n X X ≥ (1 − sk ) ak bk + s k ak b k k=1 k=1 ! 2 n X ≥ ak b k . n X

n X

(2.25)

k=1

¯ and ¯ Remark 75 Assume that ¯ a, b s are as in Corollary 74. The following inequalities hold (see [2, p. 15]). a) If

Pn

k=1

ak bk = 0, then n X

a2k

Pn

k=1

n X

b2k ≥ 4

k=1

k=1

b) If

n X

n X

!2 s k ak b k

.

(2.26)

k=1

sk ak bk = 0, then

a2k

k=1

n X k=1

 n X ak b k + b2k ≥  k=1

n X

αk a2k

k=1

n X

! 21 2 αk b2k  .

(2.27)

k=1

In particular, we may obtain the following particular inequalities involving trigonometric functions (see [2, p. 15]) n X k=1

a2k

n X

b2k

(2.28)

k=1

 n X 2  ≥ ak bk cos αk + k=1

n X k=1

a2k sin 2 αk

n X k=1

! 12 2 b2k sin2 αk



2.5. A CONDITIONAL INEQUALITY PROVIDING A REFINEMENT47 #2 " n n X X ≥ ak bk cos2 αk + ak bk sin2 αk k=1 k=1 !2 n X ≥ ak b k , k=1

where ak , bk , αk ∈ R, k = 1, . . . , n. P If one would assume that nk=1 ak bk = 0, then n X

a2k

k=1

If

Pn

k=1

n X k=1

2.5

n X

n X

b2k ≥ 4

k=1

!2 ak bk sin2 αk

.

(2.29)

k=1

ak bk sin2 αk = 0, then

a2k

n X k=1

 n X b2k ≥  ak b k +

n X

k=1

a2k sin2 αk

n X

! 21 2 b2k sin2 αk

 .

(2.30)

k=1

k=1

A Conditional Inequality Providing a Refinement

The following lemma holds [2, Lemma 4.1].

Lemma 76 Consider the sequences of real numbers x ¯ = (x1 , . . . , xn ) , y ¯= (y1 , . . . , yn ) and ¯ z = (z1 , . . . , zn ) . If yk2 ≤ |xk zk | for any k ∈ {1, . . . , n} ,

(2.31)

then one has the inequality: n X k=1

!2 |yk |

≤

n X k=1

|xk |

n X k=1

|zk | .

(2.32)

48

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

Proof. We will follow the proof in [2]. Using the condition (2.31) and the (CBS) −inequality, we have n X

|yk | ≤

k=1

n X

" 1 2

1 2

|xk | |zk | ≤

k=1

=

n X k=1

n  X

|xk |

k=1

|xk |

n X

1 2

n  2 X

|zk |

1 2

2

# 12

k=1

! 12 |zk |

k=1

which is clearly equivalent to (2.32). The following result holds [2, Theorem 4.6]. ¯ = (b1 , . . . , bn ) and ¯ Theorem 77 Let ¯ a = (a1 , . . . , an ), b c = (c1 , . . . , cn ) be sequences of real numbers such that (i) |bk | + |ck | = 6 0 (k ∈ {1, . . . , n}) (ii) |ak | ≤

2|bk ck | |bk |+|ck |

for any k ∈ {1, . . . , n} .

Then one has the inequality n X k=1

|ak | ≤

2

Pn P |bk | nk=1 |ck | k=1 Pn . k=1 (|bk | + |ck |)

(2.33)

Proof. We will follow the proof in [2]. By (ii) we observe that   2 |bk | 2 |bk ck | for any k ∈ {1, . . . , n} |ak | ≤ ≤ |bk | + |ck |  2 |ck | and thus xk := 2 |bk | − |ak | ≥ 0 zk := 2 |ck | − |ak | ≥ 0

and for any k ∈ {1, . . . , n} .

(2.34)

A simple calculation also shows that the relation (ii) is equivalent to a2k ≤ (2 |bk | − |ak |) (2 |ck | − |ak |) for any k ∈ {1, . . . , n} .

(2.35)

2.5. A CONDITIONAL INEQUALITY PROVIDING A REFINEMENT49 If we consider yk := ak and take xk , zk (k = 1, . . . , n) as defined by (2.34), then we get yk2 ≤ xk zk (with xk , zk ≥ 0) for any k ∈ {1, . . . , n}. Applying Lemma 76 we deduce ! !2 ! n n n n n X X X X X |ck | − |ak | (2.36) |ak | 2 |ak | ≤ 2 |bk | − k=1

k=1

k=1

k=1

k=1

which is clearly equivalent to (2.33). The following corollary is a natural consequence of the above theorem [2, Corollary 4.7]. Corollary 78 For any sequence x ¯ and y ¯ of real numbers, with |xk |+|yk | = 6 0 (k = 1, . . . , n) , one has: n X k=1

P P 2 nk=1 |xk | nk=1 |yk | |xk yk | ≤ Pn . |xk | + |yk | k=1 (|xk | + |yk |)

(2.37)

For two positive real numbers, let us recall the following means a+b 2 √ G (a, b) := ab

A (a, b) :=

and H (a, b) :=

1 a

2 +

(the arithmetic mean) (the geometric mean)

1 b

(the harmonic mean).

¯ = (b1 , . . . , bn ) are sequences of real We remark that if ¯ a = (a1 , . . . , an ), b numbers, then obviously ! n n n X X X A (ai , bi ) = A ai , bi , (2.38) i=1

i=1

i=1

n X

n X

and, by the (CBS) −inequality, n X i=1

G (ai , bi ) ≤ G

i=1

ai ,

! bi .

(2.39)

i=1

The following similar result for harmonic means also holds [2, p. 19].

50

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

¯ we Theorem 79 For any two sequences of positive real numbers ¯ a and b have the property: ! n n n X X X H (ai , bi ) ≤ H ai , bi . (2.40) i=1

i=1

i=1

Proof. Follows by Corollary 78 on choosing xk = ak , yk = bk and multiplying the inequality (2.37) with 2. The following refinement of the (CBS) −inequality holds [2, Corollary 4.9]. This result is known in the literature as Milne’s inequality [8]. Theorem 80 For any two sequences of real numbers p ¯ = (p1 , . . . , pn ), q ¯= (q1 , . . . , qn ) with |pk | + |qk | = 6 0 (k = 1, . . . , n) , one has the inequality: !2 n n n n n X X X X
X p2k qk2 2 2 2 p k qk ≤ p k + qk ≤ pk qk2 . (2.41) 2 2 p + q k k=1 k=1 k=1 k k=1 k=1 Proof. We shall follow the proof in [2]. The first inequality is obvious by p2 q 2 Lemma 76 on choosing yk = pk qk , xk = p2k +qk2 and zk = p2k+qk2 (k = 1, . . . , n) . k k The second inequality follows by Corollary 78 on choosing xk = p2k and y = qk2 (k = 1, . . . , n) . Remark 81 The following particular inequality is obvious by (2.41) !2 n n X X sin αk cos αk ≤n sin2 αk cos2 αk (2.42) i=1

≤

i=1 n X

2

sin αk

i=1

n X

cos2 αk ;

i=1

for any αk ∈ R, k ∈ {1, . . . , n} .

2.6

A Refinement for Non-Constant Sequences

The following result was proved in [3, Theorem 1]. ¯ = (bi ) , p Theorem 82 Let ¯ a = (ai )i∈N , b i∈N ¯ = (pi )i∈N be sequences of real numbers such that

2.6. A REFINEMENT FOR NON-CONSTANT SEQUENCES

51

(i) ai 6= aj and bi 6= bj for i 6= j, i, j ∈ N; (ii) pi > 0 for all i ∈ N. Then for any H a finite part of N one has the inequality: !2 X

X

pi a2i

i∈H

X

pi b2i −

≥ max {A, B} ≥ 0,

p i ai b i

(2.43)

i∈H

i∈H

where hP

i∈H

A := max

p i ai b i

P

j∈J

p j aj −

PJ

P

i∈H

pi a2i −

p i ai b i

P

j∈J

p j bj −

PJ

P

i∈H

pi b2i −

J⊆H J6=∅

2 i∈H pi ai

P

P

i∈J

P

j∈J

p i ai

p j bj

i2 (2.44)


2

and hP

i∈H

B := max J⊆H J6=∅

and PJ :=

P

j∈J

2 i∈H pi bi

P

P

i∈J

p i bi

P

j∈J

p j aj

i2 (2.45)


2

pj .

Proof. We shall follow the proof in [3]. Let J be a part of H. Define the mapping fJ : R → R given by  fJ (t) =

X i∈H

pi a2i 

 X

i∈H\J

pi b2i +

X

pi (bi + t)2 

i∈J

2

 −

X

i∈H\J

p i ai b i +

X

pi ai (bi + t) .

i∈J

Then by the (CBS) −inequality we have that fJ (t) ≥ 0 for all t ∈ R.

52

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY On the other hand we have " # X X X fJ (t) = pi a2i pi b2i + 2t pi bi + t2 PJ i∈H

i∈H

i∈H

#2

" −

X

p i ai b i + t

X

i∈H

p i ai

i∈J

!2 

 = t2 PJ

X

pi a2i −

i∈H

X

p i ai



i∈J

"

# X

+ 2t

pi a2i

i∈H

X

p i bi −

i∈J

p i ai b i

X

i∈H

 X X + pi a2i pi b2i − i∈H

X

i∈H

p i ai

i∈J

!2  X

p i ai b i



i∈H

for all t ∈ R. Since !2 PJ

X

pi a2i −

i∈H

X

p i ai

!2 ≥ PJ

i∈J

X

X

pi a2i −

i∈J

p i ai

>0

i∈J

as ai 6= aj for all i, j ∈ {1, . . . , n} with i 6= j, then, by the inequality fJ (t) ≥ 0 for any t ∈ R we get that " #2 X X X X 1 p i ai b i p j aj − pi a2i p j bj 0≥ ∆= 4 i∈H j∈J i∈H j∈J  !2   X X X X − PJ pi a2i − p i ai   pi a2i pi b2i − i∈H

i∈J

i∈H

i∈H

from where results the inequality !2 X i∈H

pi a2i

X i∈H

pi b2i −

X i∈H

p i ai b i

≥ A.

!2  X i∈H

p i ai b i



2.6. A REFINEMENT FOR NON-CONSTANT SEQUENCES

53

The second part of the proof goes likewise for the mapping GJ : R → R given by  gJ (t) = 

 X

pi a2i +

X

pi (ai + t)2 

i∈J

i∈H\J

X

pi b2i

i∈H

2

 −

X

p i ai b i +

X

pi bi (ai + t)

i∈J

i∈H\J

and we omit the details. The following corollary also holds [3, Corollary 1]. Corollary 83 With the assumptions of Theorem 82 and if 2 P 2 p a p b i i i i i∈H i∈H ,
2 P P 2 PH i∈H pi ai − i∈H pi ai

P

i∈H pi ai bi

C :=

P

i∈H

D :=

P

i∈H pi ai −

p i ai b i

P

i∈H

p i bi −

PH

P

i∈H

pi b2i −

P

P

i∈H

P

pi b2i

i∈H

P

pi bi

i∈H


2

p i ai

(2.46)

2 ,

(2.47)

then one has the inequality !2 X

pi a2i

X

i∈H

pi b2i −

i∈H

X

p i ai b i

≥ max {C, D} ≥ 0.

(2.48)

i∈H

The following corollary also holds [3, Corollary 2]. Corollary 84 If ai , bi 6= 0 for i ∈ N and H is a finite part of N, then one has the inequality !2 X i∈H

pi a2i

X i∈H

pi b2i −

X

pi ai bi

i∈H

1 ≥ max card (H) − 1

(P

) P 2 2 p c p d j j j j j∈H Pj∈H ,P ≥ 0, (2.49) 2 2 p a p b i∈H i i i∈H i i

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

54 where

cj := aj

X

p i ai b i − b j

i∈H

X

pi a2i , j ∈ H

(2.50)

pi ai bi , j ∈ H.

(2.51)

i∈H

and dj := aj

X

X

pi b2i − bj

i∈H

i∈H

Proof. Choosing in Theorem 82, J = {j} , we get the inequality !2 X i∈H

pi a2i

X

pi b2i −

i∈H

X

p i ai b i

≥

i∈H

pj

p2j c2j , j∈H 2 2 2 i∈H pi ai − pj aj

P

from where we obtain ! X X X pi a2i − pj a2j  pi a2i pi b2i −

X

i∈H

i∈H

i∈H

i∈H

!2  p i ai b i



≥ pj c2j for any j ∈ H. Summing these inequalities over j ∈ H, we get

[card (H) − 1]

X i∈H

 X X pi a2i  pi a2i pi b2i − i∈H

i∈H

!2  X

p i ai b i

i∈H

≥

X

pj c2j

j∈H

from where we get the first part of (2.49). The second part goes likewise and we omit the details.

Remark 85 The following particular inequalities provide refinement for the (CBS) −inequality [3, p. 60 – p. 61].

1. Assume that ¯ a = (a1 , . . . , an ), b = (b1 , . . . , bn ) are nonconstant se-

2.7. DE BRUIJN’S INEQUALITY

55

quences of real numbers. Then n X

a2i

n X

b2i −

n X

!2 ai bi

i=1    [Pn a2 Pn b − Pn a Pn a b ]2 i=1 i i=1 i i=1 i i=1 i i ≥ max , n P P  2 n 2  n ai − ( i=1 ai )  i=1   Pn Pn Pn Pn 2 2  [ i=1 bi i=1 ai bi − i=1 ai i=1 bi ]  . (2.52) n P P  2  n b2i − ( ni=1 bi ) 

i=1

i=1

i=1

¯ are sequences of real numbers with not all elements 2. Assume that ¯ a and b equal to zero, then n X i=1

a2i

n X

b2i −

i=1

n X

!2 ai b i

i=1

≥

  2 n n n P P P  2  aj ai b i − b j ai  

1 j=1 max  n−1   

i=1

i=1

n P i=1 n P j=1

 aj

n P

a2i

− bj

i=1

n P i=1

n P i=1

2.7

,

a2i

b2i

2   ai b i   

. (2.53)

   

De Bruijn’s Inequality

The following refinement of the (CBS) −inequality was proved by N.G. de Bruijn in 1960, [4] (see also [5, p. 89]).

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

56

Theorem 86 If ¯ a = (a1 , . . . , an ) is a sequence of real numbers and ¯ z = (z1 , . . . , zn ) is a sequence of complex numbers, then 2 # " n n n n X X X X 1 2 2 2 ak zk ≤ a |zk | + zk . (2.54) 2 k=1 k k=1 k=1 k=1 Equality holds in (2.54) if and only ifP for k ∈ {1, . . . , n} , ak = Re (λzk ) , where λ is a complex number such that nk=1 λ2 zk2 is a nonnegative real number. Proof. We shall follow the proof in [5, p. 89 – p. 90]. By a simultaneous rotation of all the zk ’s about the origin, we get n X

ak zk ≥ 0.

k=1

This rotation does not affect the moduli n n X X 2 a z , z k k k and |zk | for k ∈ {1, . . . , n} . k=1 k=1 P Hence, it is sufficient to prove inequality (2.54) for the case where nk=1 ak zk ≥ 0. If we put zk = xk + iyk (k ∈ {1, . . . , n}) , then, by the (CBS) −inequality for real numbers, we have 2 !2 n n n n X X X X 2 ak zk = ak zk ≤ ak x2k . (2.55) k=1

k=1

k=1

k=1

Since 2x2k = |zk |2 + Re zk2 for any k ∈ {1, . . . , n} we obtain, by (2.55), that n 2 " n # n n X X 1X 2 X ak |zk |2 + Re zk2 . ak zk ≤ 2 k=1 k=1 k=1 k=1 As

n X k=1

Re zk2 = Re

n X k=1

! zk2

n X 2 ≤ zk , k=1

then by (2.56) we deduce the desired inequality (2.54).

(2.56)

2.8. MCLAUGHLIN’S INEQUALITY

2.8

57

McLaughlin’s Inequality

The following refinement of the (CBS) −inequality for sequences of real numbers was obtained in 1966 by H.W. McLaughlin [7, p. 66]. ¯ = (b1 , . . . , b2n ) are sequences of real Theorem 87 If ¯ a = (a1 , . . . , a2n ), b numbers, then 2n X

!2 ai b i

" +

i=1

n X

#2 (ai bn+i − an+i bi )

i=1

≤

2n X

a2i

2n X

i=1

b2i

(2.57)

i=1

with equality if and only if for any i, j ∈ {1, . . . , n} ai bj − aj bi − an+i bn+j + an+j bn+i = 0

(2.58)

ai bn+j − aj bn+i + an+i bj − an+j bi = 0.

(2.59)

and

Proof. We shall follow the proof in [6] by M.O. Drˆimbe. The following identity may be obtained by direct computation 2n X i=1

a2i

2n X

b2i −

i=1

2n X

!2 ai b i

" −

#2 (ai bn+i − an+i bi )

i=1

i=1

=

n X

X

(ai bj − aj bi − an+i bn+j + an+j bn+i )2

1≤i<j≤n

+

X

(ai bn+j − aj bn+i + an+i bj − an+j bi )2 . (2.60)

1≤i<j≤n

It is obvious that (2.57) is a simple consequence of the identity (2.60). The case of equality is also obvious. Remark 88 For other similar (CBS) −type inequalties see the survey paper ¯ having 4n terms [7]. An analogous inequality to (2.57) for sequences ¯ a and b each may be found in [7, p. 70].

58

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

2.9

A Refinement due to Daykin-Eliezer-Carlitz

We will present now the version due to Mitrinovi´c, Peˇcari´c and Fink [5, p. 87] of Daykin-Eliezer-Carlitz’s refinement of the discrete (CBS) −inequality [8]. ¯ = (b1 , . . . , bn ) be two sequences of Theorem 89 Let ¯ a = (a1 , . . . , an ) and b positive numbers. The ineuality !2 n n n n n X X X X X 2 ai b i ≤ f (ai bi ) g (ai bi ) ≤ (2.61) ai b2i i=1

i=1

i=1

i=1

i=1

holds if and only if f (a, b) g (a, b) = a2 b2 , f (ka, kb) = k 2 f (a, b) , a b bf (a, 1) af (b, 1) + ≤ + af (b, 1) bf (a, 1) b a

(2.62) (2.63) (2.64)

for any a, b, k > 0. Proof. We shall follow the proof in [5, p. 88 – p. 89]. Necessity. Indeed, for n = 1, the inequality (2.61) becomes (ab)2 ≤ f (a, b) g (a, b) ≤ a2 b2 ,

a, b > 0

which gives the condition (2.62). For n = 2 in (2.61), using (2.62), we get 2a1 b1 a2 b2 ≤ f (a1 , b1 ) g (a2 , b2 ) + f (a2 , b2 ) g (a1 , b1 ) ≤ a21 b22 + a22 b21 . By eliminating g, we get 2≤

f (a1 , b1 ) a2 b2 f (a2 , b2 ) a1 b1 a1 b 2 a2 b 1 · + · ≤ + . f (a2 , b2 ) a1 b1 f (a1 , b1 ) a2 b2 a2 b 1 a1 b 2

(2.65)

By substituting in (2.65) a, b for a1 , b1 and ka, kb for a2 , b2 (k > 0), we get 2≤

f (a, b) 2 f (ka, kb) −2 k + k ≤2 f (ka, kb) f (a, b)

2.9. A REFINEMENT DUE TO DAYKIN-ELIEZER-CARLITZ

59

and this is valid only if k 2 f (a, b) (f (ka, kb)) = 1, i.e., the condition (2.63) holds. Using (2.65), for a1 = a, b1 = b, a2 = b, b2 = 1, we have 2≤

f (a,1) a f (b,1) b

+

f (b,1) b f (a,1) a

≤

a b + . b a

(2.66)

The first inequality in (2.66) is always satisfied while the second inequality is equivalent to (2.64). Sufficiency. Suppose that (2.62) holds. Then inequality (2.61) can be written in the form X X 2 ai b i aj b j ≤ [f (ai , bi ) g (aj , bj ) + f (aj , bj ) g (ai , bi )] 1≤i<j≤n

1≤i<j≤n

X

≤


a2i b2j + a2j b2i .

1≤i<j≤n

Therefore, it is enough to prove 2ai bi aj bj ≤ f (ai , bi ) g (aj , bj ) + f (aj , bj ) g (ai , bi ) ≤ a2i b2j + a2j b2i . Suppose that (2.64) holds. Then (2.66) holds and putting a = (2.66) and using (2.63), we get 2≤

(2.67) ai , bi

b=

aj bj

in

f (ai , bi ) aj bj f (aj , bj ) ai bi ai b j aj b i · + · ≤ + . f (aj , bj ) ai bi f (ai , bi ) aj bj aj bi ai bj

Multiplying the last inequality by ai bi aj bj and using (2.62), we obtain (2.67). Remark 90 In [8] (see [5, p. 89]) the condition (2.64) is given as f (a, 1) f (b, 1) ≤ for a ≥ b > 0. (2.68) 2 a b2 Remark 91 O.E. Daykin, C.J. Eliezer and C. Carlitz [8] stated that examples for f, g satisfying (2.62) – (2.64) were obtained in the literature. The 2 y2 choice f (x, y) = x2 + y 2 , g (x, y) = xx2 +y 2 will give the Milne’s inequality !2 n n n n n X X X X
X a2i b2i 2 ai b i ≤ a2i + b2i · ≤ a · b2i . (2.69) i 2 2 a + b i i i=1 i=1 i=1 i=1 i=1 f (b, 1) ≤ f (a, 1) ,

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

60

For a different proof of this fact, see Section 2.5. The choice f (x, y) = x1+α y 1−α , g (x, y) = x1−α y 1+α (α ∈ [0, 1]) will give the Callebaut inequality !2 n n n n n X X X X X 1+α 1−α 1−α 1+α 2 (2.70) ai b i ≤ ai b i ai b i ≤ ai · b2i . i=1

2.10

i=1

i=1

i=1

i=1

A Refinement via Dunkl-Williams’ Inequality

We will use the following version of Dunkl-Williams’ inequality established in 1964 in inner product spaces [9]. Lemma 92 Let a, b be two non-null complex numbers. Then a b 1 |a − b| ≥ (|a| + |b|) − . 2 |a| |b|

(2.71)

Proof. We start with the identity (see also [5, pp. 515 – 516]) 2   ¯b  a a ¯ − b = a − b − |a| |b| |a| |b| |a| |b|   a ¯b · = 2 − 2 Re |a| |b|

1 = 2 |a| |b| − 2 Re a · ¯b |a| |b|
 1  = 2 |a| |b| − |a|2 + |b|2 − |a − b|2 |a| |b|  1  = |a − b|2 − (|a| − |b|)2 . |a| |b| Hence 2 2 a 1 b |a − b| − (|a| + |b|) − 2 |a| |b| 2



=

 (|a| − |b|)2  (|a| + |b|)2 − |a − b|2 4 |a| |b|

2.11. SOME REFINEMENTS DUE TO ALZER AND ZHENG

61

and (2.71) is proved. Using the above result, we may prove the following refinement of the (CBS) −inequality for complex numbers. ¯ = (b1 , . . . , bn ) are two sequences of Theorem 93 If ¯ a = (a1 , . . . , an ), b nonzero complex numbers, then 2 n n n X X X 2 2 |ak | |bk | − ak b k k=1 k=1 k=1 2 n |bi | |aj | |ai | |bj | 1 X a ¯ i bj − a ¯ j bi + a ¯i · bj − bi · a ¯j ≥ 0 (2.72) ≥ 8 i,j=1 |ai | |bj | |bi | |aj | Proof. The inequality (2.71) is clearly equivalent to 2 |b| |a| 1 2 ·a− ·b (2.73) |a − b| ≥ a − b + 4 |a| |b| for any a, b ∈ C, a, b 6= 0. We know the Lagrange’s identity for sequences of complex numbers (see Chapter I, Section 1.2 ) 2 n n n n X X X 1X |ak |2 |bk |2 − ak b k = |¯ ai b j − a ¯j bi |2 . (2.74) 2 i,j=1 k=1 k=1 k=1 By (2.73), we have 2 1 |aj | |bi | |ai | |bj | |¯ ai b j − a ¯ j bi | ≥ a ¯ i bj − a ¯ j bi + a ¯ i bj − a ¯j bi . 4 |ai | |bj | |aj | |bi | 2

Summing over i, j from 1 to n and using the (CBS) −inequality for double sums, we deduce (2.72).

2.11

Some Refinements due to Alzer and Zheng

In 1992, H. Alzer [10] presented the following refinement of the CauchySchwartz inequality written in the form !2 n n n X X X xk yk ≤ yk x2k yk . (2.75) k=1

k=1

k=1

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

62

Theorem 94 Let xk and yk (k = 1, . . . , n) be real numbers satisfying 0 = x0 < x1 ≤ x22 ≤ · · · ≤ xnn and 0 < yn ≤ yn−1 ≤ · · · ≤ y1 . Then n X

!2

 n  X 1 2 ≤ yk xk − xk−1 xk yk , 4 k=1 k=1 n X

xk yk

k=1

(2.76)

with equality holding if and only if xk = kx1 (k = 1, . . . , n) and y1 = · · · = yn . In 1998, Liu Zheng [11] pointed out an error in the proof given in [10], which can be corrected as shown in [11]. Moreover, Liu Zheng established the following result which sharpens (2.76). Theorem 95 Let xk and yk (k = 1, . . . , n) be real numbers satisfying 0 < x1 ≤ x22 ≤ · · · ≤ xnn and 0 < yn ≤ yn−1 ≤ · · · ≤ y1 . Then n X

!2 ≤

xk yk

k=1

n X k=1

yk

n X

δ k yk ,

(2.77)

k=1

with δ 1 = x21 and δ k =

k 7k + 1 2 xk − x2 8k 8 (k − 1) k−1

(k ≥ 2) .

(2.78)

Equality holds in (2.77) if and only if xk = kx1 (k = 1, . . . , n) and y1 = · · · = yn . In 1999, H. Alzer improved the above results as follows. To present his results, we will follow [12]. In order to prove the main result, we need some technical lemmas. Lemma 96 Let xk (k = 1, . . . , n) be real numbers such that 0 < x1 ≤ Then 2

n X

x2 xn ≤ ··· ≤ . 2 n

xk ≤ (n + 1) xn ,

k=1

with equality holding if and only if xk = kx1 (k = 1, . . . , n) .

(2.79)

2.11. SOME REFINEMENTS DUE TO ALZER AND ZHENG

63

A proof of Lemma 96 is given in [10]. Lemma 97 Let xk (k = 1, . . . , n) be real numbers such that 0 < x1 ≤

x2 xn ≤ ··· ≤ . 2 n

Then n X

!2 ≤n

xk

k=1

n X 3k + 1

4k

k=1

x2k ,

(2.80)

with equality holding if and only if xk = kx1 (k = 1, . . . , n) . Proof. Let Sn = Sn (x1 , . . . , xn ) = n

n X 3k + 1

4k

k=1

x2k −

n X

!2 xk

.

k=1

Then we have for n ≥ 2 : Sn − Sn−1 =

n−1 X 3k + 1 k=1

4k

x2k − 2xn

n−1 X k=1

xk +

3 (n − 1) 2 xn 4

(2.81)

= f (xn ) , say. We differentiate with respect to xn and use (2.79) and xn ≥ yields n−1 X 3 (n − 1) n 0 f (xn ) = xn − 2 xk ≥ xn−1 > 0 2 2 k=1

n x . n−1 n−1

This

and  f (xn ) ≥ f =

n xn−1 n−1

n X 3k + 1 k=1

4k

x2k

 (2.82)

n−1 X 2n 3n2 − xn−1 xk + x2n−1 n−1 4 (n − 1) k=1

= Tn−1 (x1 , . . . , xn−1 ) ,

say.

64

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

We use induction on n to establish that Tn−1 (x1 , . . . , xn−1 ) ≥ 0 for n ≥ 2. We have T1 (x1 ) = 0. Let n ≥ 3; applying (2.79) we obtain n−1

∂ 3n + 2 2n X xk Tn−1 (x1 , . . . , xn−1 ) = xn−1 − ∂xn−1 2 n − 1 k=1 ≥

(n − 2) (n + 1) xn−1 > 0 2 (n − 1)

and  Tn−1 (x1 , . . . , xn−1 ) ≥ Tn−1

 n−1 x1 , . . . , xn−2 , xn−2 . n−2

(2.83)

Using the induction hypothesis Tn−2 (x1 , . . . , xn−2 ) ≥ 0 and (2.79), we get " #   n−2 X n−1 xn−2 Tn−1 x1 , . . . , xn−2 , xn−2 ≥ (n − 1) xn−2 − 2 xk . (2.84) n−2 n−2 k=1 From (2.83) and (2.84) we conclude Tn−1 (x1 , . . . , xn−1 ) ≥ 0 for n ≥ 2, so that (2.81) and (2.82) imply Sn ≥ Sn−1 ≥ · · · ≥ S2 ≥ S1 = 0.

(2.85)

This proves inequality (2.80). We discuss the cases of equality. A simple calculation reveals that Sn (x1 , 2x1 , . . . , nx1 ) = 0. We use induction on n to prove the implication Sn (x1 , . . . , xn ) = 0 =⇒ xk = kx1 for k = 1, . . . , n.

(2.86)

If n = 1, then (2.86) is obviously true. Next, we assume that (2.86) holds with n − 1 instead of n. Let n ≥ 2 and Sn (x1 , . . . , xn ) = 0. Then (2.85) leads to Sn−1 (x1 , . . . , xn−1 ) = 0 which implies xk = kx1 for k = 1, . . . , n − 1. Thus, we have Sn (x1 , 2x1 , . . . , (n − 1) x1 , xn ) = 0 which is equivalent to (xn − nx1 ) (3xn − nx1 ) = 0. Since 3xn > nx1 , we get xn = nx1 . Lemma 98 Let xk (k = 1, . . . , n) be real numbers such that x2 xn 0 < x1 ≤ ≤ ··· ≤ . 2 n If the natural numbers n and q satisfy n ≥ q + 1, then !2 q q X X (3n + 1) q 2 2 xn . 0< xk − 2qxn xk + 4n k=1 k=1

(2.87)

2.11. SOME REFINEMENTS DUE TO ALZER AND ZHENG

65

Proof. We denote the expression on the right-hand side of (2.87) by u (xn ) . Then we differentiate with respect to xn and apply (2.79), xn ≥  n xq and n ≥ q + 1. This yields q q X 1 0 (3n + 1) q u (xn ) = xn − 2q xk q 2n k=1

(3n + 1) q xn − (q + 1) xq 2n 3n − 2q − 1 ≥ xq 2 > 0.

≥

Hence, we get  u (xn ) ≥ u Let

n xq q



q

X (3n + 1) n 2 = xq − 2nxq xk + 4 k=1

q X

!2 xk

.

(2.88)

k=1

q X (3t + 1) t xq − 2t xk and t ≥ q + 1; v (t) = 4 k=1

from (2.79) we conclude that q X 6t + 1 2q + 3 v (t) = xq − 2 xk ≥ xq > 0. 4 4 k=1 0

This implies that the expression on the right-hand side of (2.88) is increasing on [q + 1, ∞) with respect to n. Since n ≥ q + 1, we get from (2.88): !2 q q X X (3q + 4) (q + 1) 2 xq − 2 (q + 1) xq xk + xk (2.89) u (xn ) ≥ 4 k=1 k=1 = Pq (x1 , . . . , xq ) ,

say.

We use induction on q to show that Pq (x1 , . . . , xq ) > 0 for q ≥ 1. We have P1 (x1 ) = 21 x21 . If Pq−1 (x1 , . . . , xq−1 ) > 0, then we obtain for q ≥ 2 : Pq (x1 , . . . , xq ) > 2q (xq−1 − xq )

q−1 X

xk −

k=1

+

(3q + 1) q 2 xq−1 4

q (3q − 1) 2 xq = w (xq ) , say. (2.90) 4

66

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

We differentiate with respect to xq and use (2.79) and xq ≥ we get # " q−1 X 3q − 1 w0 (xq ) = q xq − 2 xk 2 k=1



q q−1



xq−1 . Then

q 2 (q + 1) ≥ xq−1 2 (q − 1) >0 and 

 q w (xq ) ≥ w xq−1 q−1 " # q−1 X 4q 2 − q − 1 q xq−1 xq−1 − 2 xk = q−1 4 (q − 1) k=1

(2.91)

(3q − 1) q 2 x 4 (q − 1) q−1 > 0.

≥

From (2.89), (2.90) and (2.91), we obtain u (xn ) > 0. We are now in a position to prove the following companion of inequalities (2.76) and (2.77) (see [12]). Theorem 99 The inequality n X

!2 ≤

xk yk

k=1

n X k=1

yk

n  X k=1

β α+ k

 yk ,

(2.92)

holds for all natural numbers n and for all real numbers xk and yk (k = 1, . . . , n) with 0 < x1 ≤

x2 xn ≤ ··· ≤ and 0 < yn ≤ yn−1 ≤ · · · ≤ y1 , 2 n

if and only if α≥

3 4

and β ≥ 1 − α.

(2.93)

2.11. SOME REFINEMENTS DUE TO ALZER AND ZHENG

67

Proof. First, we assume that (2.92) is valid for all n ≥ 1 and for all real numbers xk and yk (k = 1, . . . , n) which satisfy (2.93). We set xk = k and yk = 1 (k = 1, . . . , n). Then (2.92) leads to  0≤

3 α− 4

 2n + α + 3β −

3 2

(n ≥ 1) .

(2.94)

This implies α ≥ 34 . And, (2.94) with n = 1 yields α + β ≥ 1. Now, we suppose that α ≥ 43 and β ≥ 1 − α. Then we obtain for k ≥ 1 : α+

β 1−α 3 1 ≥α+ ≥ + , k k 4 4k

so that is suffices to show that (2.92) holds with α =

F (y1 , . . . , yn ) =

n X k=1

yk

n X 3k + 1 k=1

4k

x2k yk −

3 4

and β = 14 . Let

n X

!2 xk yk

k=1

and Fq (y) = F (y, . . . , y, yq+1 , . . . , yn )

(1 ≤ q ≤ n − 1) .

We shall prove that Fq is strictly increasing on [yq+1 , ∞). Since yq+1 ≤ yq , we obtain Fq (yq ) ≥ Fq (yq+1 ) = Fq+1 (yq+1 )

(1 ≤ q ≤ n − 1) ,

(2.95)

and Lemma 97 imply F (y1 , . . . , yn ) = F1 (y1 ) ≥ F1 (y2 ) = F2 (y2 ) ≥ F2 (y3 ) ≥ · · · ≥ Fn−1 (yn−1 ) ≥ Fn−1 (y)  !2  n n X X 3k + 1 2 = yn2 n xk − xk  ≥ 0. 4k k=1 k=1 If F (y1 , . . . , yn ) = 0, then we conclude from the strict monotonicity of Fq and from Lemma 97 that y1 = · · · = yn and xk = kx1 (k = 1, . . . , n) .

68

CHAPTER 2. REFINEMENTS OF THE (CBS) −INEQUALITY

It remains to show that Fq is strictly increasing on [yq+1 , ∞). Let y ≥ yq+1 ; we differentiate Fq and apply Lemma 97. This yields  !2  q q n X X X 3k + 1 2 3k + 1 2 0   Fq (y) = 2y q xk − xk +q xk yk 4k 4k k=1 k=q+1 k=1 +

n X k=q+1

yk

q X 3k + 1 k=1

4k

x2k − 2

q X

xk yk

k=q+1

and q X 1 00 3k + 1 2 Fq (y) = q xk − 2 4k k=1

n X

q X

xk

k=1

!2 xk

≥ 0.

k=1

Hence, we have Fq0 (y) ≥ Fq0 (yq+1 )

(2.96)

!2  ! q q n X X X 3k + 1 1 = 2qyq+1 + yk  x2k − xk  4k q k=1 k=q+1 k=1  !2  q q n   2 X X 1 X (3k + 1) q 2 + yk xk − 2qxk xi + xi .  q k=q+1  4k i=1 i=1 From Lemma 97 and Lemma 98 we obtain Fq0 (yq+1 ) > 0, so that (2.96) implies Fq0 (y) > 0 for y ≥ yq+1 . This completes the proof of the theorem. Remark 100 The proof of the theorem reveals that the sign of equality holds in (2.92) (with α = 34 and β = 14 ) if and only if xk = kx1 (k = 1, . . . , n) and y1 = · · · = yn . Remark 101 If δ k is defined by (2.78), then we have for k ≥ 2 : "   2 #  x 2  x 3 1 k (k − 1) k k−1 δk − + x2k = − , 4 4k 8 k k−1 which implies that inequality (2.92) (with α =

3 4

and β = 14 ) sharpens (2.77).

Remark 102 It is shown in [10] that
if a sequence (xk ) satisfies x0 = 0 and xk 2xk ≤ xk−1 + xk+1 (k ≥ 1) , then k is increasing. Hence, inequality (2.92) is valid for all sequences (xk ) which are positive and convex.

Bibliography [1] S.S. DRAGOMIR and N.M. IONESCU, Some refinements of CauchyBuniakowski-Schwartz inequality for sequences, Proc. of the Third Symp. of Math. and Its Appl., 3-4 Nov. 1989, Polytechnical Institute of Timi¸soara, Romania, 78-82. [2] S.S. DRAGOMIR, On Cauchy-Buniakowski-Schwartz’s inequality for real numbers (Romanian), “Caiete Metodico S¸tiint¸ifice”, No. 57, 1989, pp. 24. Faculty of Mathematics, Timi¸soara University, Romania. ˇ ´ and D.M. MILOSEVI ˇ ´ On [3] S.S. DRAGOMIR, S.Z. ARSLANAGIC C, Cauchy-Buniakowski-Schwartz’s inequality for real numbers, Mat. Bilten, 18 (1994), 57-62. [4] N.G. de BRUIJN, Problem 12, Wisk. Opgaven, 21 (1960), 12-14. ´ J.E. PECARI ˘ ´ and A.M. FINK, Classical and [5] D.S. MITRINOVIC, C New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993. [6] M.O. DRˆIMBE, A generalisation of Lagrange’s identity (Romanian), Gaz. Mat., (Bucharest), 89 (1984), 139-141. [7] H.W. McLAUGHLIN, Inequalities complementary to the CauchySchwartz inequality for finite sums of quaternions, refinements of the Cauchy-Schwartz inequality for finite sums of real numbers; inequalities concerning ratios and differences of generalised means of different order, Maryland University, Technical Note BN-454, (1966), 129 pp. [8] D.E. DAYKIN, C.J. ELIEZER and C. CARLITZ, Problem 5563, Amer. Math. Monthly, 75 (1968), p. 198 and 76 (1969), 98-100. 69

70

BIBLIOGRAPHY

[9] C.F. DUNKL and K.S. WILLIAMS, A simple inequality, Amer. Math. Monthly, 71 (1964), 53-54. [10] H. ALZER, A refinement of the Cauchy-Schwartz inequality, J. Math. Anal. Appl., 168 (1992), 596-604. [11] L. ZHENG, Remark on a refinement of the Cauchy-Schwartz inequality, J. Math. Anal. Appl., 218 (1998), 13-21. [12] H. ALZER, On the Cauchy-Schwartz inequality, J. Math. Anal. Appl., 234 (1999), 6-14.

Chapter 3 Functional Properties 3.1

A Monotonicity Property

The following result was obtained in [1, Theorem]. ¯ = (b1 , . . . , bn ) be sequences of real Theorem 103 Let ¯ a = (a1 , . . . , an ), b numbers and p ¯ = (p1 , . . . , pn ) , q ¯ = (q1 , . . . , qn ) be sequences of nonnegative real numbers such that pk ≥ qk for any k ∈ {1, . . . , n} . Then one has the inequality n X

! 21 pi a2i

i=1

n X

! 12 pi b2i

i=1

≥

n X − pi ai bi i=1 ! 12 ! 12 n n n X X X qi a2i qi b2i − qi ai bi ≥ 0. (3.1) i=1

i=1

i=1

Proof. We shall follow the proof in [1]. Since pk −qk ≥ 0, then the (CBS) −inequality for the weights rk = pk −qk (k ∈ {1, . . . , n}) will produce ! n ! " n #2 n n n X X X X X pk a2k − qk a2k pk b2k − qk b2k ≥ (pk −qk ) ak bk . (3.2) k=1

k=1

k=1

k=1

k=1

Using the elementary inequality (ac − bd)2 ≥ a2 − b2





c2 − d2 , 71

a, b, c, d ∈ R

72

CHAPTER 3. FUNCTIONAL PROPERTIES

for the choices n X

a =

! 12 pk a2k

,

b=

k=1 n X

d =

n X

! 12 qk a2k

,

n X

c=

k=1

! 12 pk b2k

and

k=1

! 21 qk b2k

k=1

we deduce by (3.2), that n X

! 12 pk a2k

k=1

! 12

n X

pk b2k

−

n X

! 12 qk a2k

n X

! 12 qk b2k

k=1

k=1 k=1 n n n n X X X X ≥ p k ak b k − q k ak b k ≥ p k ak b k − q k ak b k k=1

k=1

k=1

k=1

proving the desired inequality (3.1). The following corollary holds [1, Corollary 1]. ¯ be as in Theorem 103. Denote Corollary 104 Let ¯ a and b Sn (1) := {¯ x = (x1 , . . . , xn ) |0 ≤ xi ≤ 1, i ∈ {1, . . . , n}} . Then n X i=1

! 21 a2i

! 12

n X b2i − ai b i i=1 i=1  ! 12 n X = sup  xi a2i n X

x ¯∈Sn (1)

i=1

n X i=1

! 12 xi b2i

 n X − xi ai bi  ≥ 0. (3.3) i=1

Remark 105 The following inequality is a natural particular case that may

3.2. A SUPERADDITIVITY PROPERTY IN TERMS OF WEIGHTS 73 be obtained from (3.1) [1, p. 79] n X i=1

! 21 a2i

n X

! 12 b2i

i=1

i=1

" ≥

n X − ai b i

n X

# 12 " a2i trig2 (αi )

n X

# 12 b2i trig2 (αi )

i=1

i=1

n X 2 − ai bi trig (αi ) ≥ 0, (3.4) i=1

where trig (x) = sin x or cos x, x ∈ R and α ¯ = (α1 , . . . , αn ) is a sequence of real numbers.

3.2

A Superadditivity Property in Terms of Weights

Let Pf (N) be the family of finite parts of the set of natural numbers N, S (K) the linear space of real or complex numbers, i.e.,  S (K) := x|x = (xi )i∈N , xi ∈ K, i ∈ N and S+ (R) the family of nonnegative real sequences. Define the mapping ! 21 S (p, I, x, y) :=

X

2

pi |xi |

i∈I

X i∈I

2

pi |yi |

X − pi xi y¯i ,

(3.5)

i∈I

where p ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) . The following superadditivity property in terms of weights holds [2, p. 16]. Theorem 106 For any p, q ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) we have S (p + q, I, x, y) ≥ S (p, I, x, y) + S (q, I, x, y) ≥ 0.

(3.6)

Proof. Using the (CBS) −inequality for real numbers a2 + b 2


21

c2 + d2


12

≥ ac + bd; a, b, c, d ≥ 0,

(3.7)

74

CHAPTER 3. FUNCTIONAL PROPERTIES

we have S (p, I, x, y) ! 12 =

X i∈I

2

pi |xi | +

X

2

qi |xi |

i∈I

! 12 X

pi |yi | +

i∈I

X X − pi xi y¯i + qi xi y¯i i∈I i∈I ! 12 ! 12 X X ≥ pi |xi |2 pi |yi |2 + i∈I i∈I X X − pi xi y¯i − qi xi y¯i i∈I

2

X

2

qi |yi |

i∈I

! 12 X

qi |xi |2

i∈I

! 12 X

qi |yi |2

i∈I

i∈I

= S (p, I, x, y) + S (q, I, x, y) , and the inequality (3.6) is proved. The following corollary concerning the monotonicity of S (·, I, x, y) also holds [2, p. 16]. Corollary 107 For any p, q ∈ S+ (R) with p ≥ q and I ∈ Pf (N) , x, y ∈ S (K) one has the inequality: S (p, I, x, y) ≥ S (q, I, x, y) ≥ 0.

(3.8)

Proof. Using Theorem 106, we have S (p, I, x, y) = S ((p − q) + q, I, x, y) ≥ S (p − q, I, x, y) + S (q, I, x, y) giving S (p, I, x, y) − S (q, I, x, y) ≥ S (p − q, I, x, y) ≥ 0 and the inequality (3.8) is proved. Remark 108 The following inequalities follow by the above results [2, p. 17].

3.3. THE SUPERADDITIVITY AS AN INDEX SET MAPPING

75

1. Let αi ∈ R (i ∈ {1, . . . , n}) and xi , yi ∈ K (i ∈ {1, . . . , n}) . Then one has the inequality: n X

n X |xi |2 |yi |2

i=1

i=1

! 21

n X − xi y¯i i=1

+

! 12

n X − xi y¯i sin2 αi ≥ |xi |2 sin2 αi |yi |2 sin2 αi i=1 i=1 i=1 ! 12 n n n X X X |xi |2 cos2 αi |yi |2 cos2 αi − xi y¯i cos2 αi ≥ 0. (3.9) n X

n X

i=1

i=1

i=1

2. Denote Sn (1) := {p ∈ S+ (R) |pi ≤ 1 for all i ∈ {1, . . . , n}} . Then for all x, y ∈ S (K) one has the bound (see also Corollary 104): ! 12

n X 0≤ − xi y¯i (3.10) i=1 i=1 i=1   ! 12 n n n  X  X X 2 2 = sup pi |xi | pi |yi | − pi xi y¯i .   p∈Sn (1) n X

n X 2 |xi | |yi |2

i=1

3.3

i=1

i=1

The Superadditivity as an Index Set Mapping

We assume that we are under the hypothesis and notations in Section 3.2. Reconsider the functional S (·, ·, ·, ·) : S+ (R) × Pf (N) × S (K) × S (K) → R, ! 21 S (p, I, x, y) :=

X

X pi |xi |2 pi |yi |2

i∈I

i∈I

X − pi xi y¯i .

(3.11)

i∈I

The following superadditivity property as an index set mapping holds [2]. Theorem 109 For any I, J ∈ Pf (N) \ {∅} with I ∩ J = ∅, one has the inequality S (p, I ∪ J, x, y) ≥ S (p, I, x, y) + S (p, J, x, y) ≥ 0.

(3.12)

76

CHAPTER 3. FUNCTIONAL PROPERTIES Proof. Using the elementary inequality for real numbers a2 + b 2


12

c2 + d2


12

≥ ac + bd; a, b, c, d ≥ 0,

(3.13)

we have S (p, I ∪ J, x, y) ! 12 =

X i∈I

2

pi |xi | +

X

2

pj |xj |

j∈J

! 12 X

pi |yi | +

i∈I

X X − pi xi y¯i + pj xj y¯j i∈I j∈J ! 12 ! 12 X X ≥ pi |xi |2 pi |yi |2 + i∈I

2

X

2

pj |yj |

j∈J

! 12 X

i∈I

pj |xj |2

j∈J

! 12 X

pj |yj |2

j∈J

X X − pi xi y¯i − pj xj y¯j i∈I

j∈J

= S (p, I, x, y) + S (p, J, x, y) and the inequality (3.12) is proved. The following corollary concerning the monotonicity of S (p, ·, x, y) as an index set mapping also holds [2, p. 16]. Corollary 110 For any I, J ∈ Pf (N) with I ⊇ J 6= ∅, one has S (p, I, x, y) ≥ S (p, J, x, y) ≥ 0.

(3.14)

Proof. Using Theorem 109, we may write S (p, I, x, y) = S (p, (I\J) ∪ J, x, y) ≥ S (p, I\J, x, y) + S (p, J, x, y) giving S (p, I, x, y) − S (p, J, x, y) ≥ S (p, I\J, x, y) ≥ 0 which proves the desired inequality (3.14). Remark 111 The following inequalities follow by the above results [2, p. 17].

3.3. THE SUPERADDITIVITY AS AN INDEX SET MAPPING

77

1. Let pi ≥ 0 (i ∈ {1, . . . , 2n}) and xi , yi ∈ K (i ∈ {1, . . . , 2n}) . Then we have the inequality ! 12

2n X − pi xi y¯i (3.15) i=1 i=1 i=1 1 ! n n n 2 X X X p2i |x2i |2 p2i |y2i |2 − p2i xi y¯i ≥ i=1 i=1 i=1 ! 12 n n n X X X + p2i−1 |x2i−1 |2 p2i−1 |y2i−1 |2 − p2i−1 x2i−1 y¯2i−1 2n X

2n X pi |xi |2 pi |yi |2

i=1

i=1

i=1

≥ 0. 2. We have the bound ! 12

n X − pi xi y¯i i=1 i=1 i=1   ! 21 X X X = sup  − pi xi y¯i  ≥ 0. (3.16) pi |xi |2 pi |yi |2 I⊆{1,...,n}

n X

n X 2 pi |xi | pi |yi |2

i∈I

I6=∅

i∈I

i∈I

3. Define the sequence

Sn :=

n X i=1

2

pi |xi |

n X i=1

! 12 2

pi |yi |

n X − pi xi y¯i ≥ 0

(3.17)

i=1

where p = (pi )i∈N ∈ S+ (R) , x = (xi )i∈N , y = (yi )i∈N ∈ S (K) . Then Sn is monotontic nondecreasing and we have the following lower bound n
1
1 (3.18) Sn ≥ max pi |xi |2 + pj |xj |2 2 pi |yi |2 + pj |yj |2 2 1≤i,j≤n  − |pi xi y¯i + pj xj y¯j | ≥ 0.

78

CHAPTER 3. FUNCTIONAL PROPERTIES

3.4

Strong Superadditivity in Terms of Weights

With the notations in Section 3.2, define the mapping 2 X X X 2 2 ¯ S (p, I, x, y) := pi |xi | pi |yi | − pi xi y¯i , i∈I

i∈I

(3.19)

i∈I

where p ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) . Denote also by k·k`,H the weighted Euclidean norm ! 21 X

kxk`,H :=

`i |xi |2

,

` ∈ S+ (R) , H ∈ Pf (N) .

(3.20)

i∈H

The following strong superadditivity property in terms of weights holds [2, p. 18]. Theorem 112 For any p, q ∈ S+ (R) , I ∈ Pf (N) and x, y ∈ S (K) we have S¯ (p + q, I, x, y) − S¯ (p, I, x, y) − S¯ (q, I, x, y)   kxkp,I   ≥ det kxkq,I

kykp,I

2  ≥ 0. (3.21)

kykq,I

Proof. We have S¯ (p + q, I, x, y) ! ! X X X X = pi |xi |2 + qi |xi |2 pi |yi |2 + qi |yi |2 i∈I

i∈I

i∈I

i∈I

2 X X − pi xi y¯i + qi xi y¯i i∈I i∈I X X X X ≥ pi |xi |2 pi |yi |2 + qi |xi |2 qi |yi |2 i∈I

−

i∈I

i∈I

!2 X X pi xi y¯i + qi xi y¯i i∈I

i∈I

i∈I

(3.22)

3.4. STRONG SUPERADDITIVITY IN TERMS OF WEIGHTS = S¯ (p, I, x, y) + S¯ (q, I, x, y) +

X

pi |xi |2

X

79

qi |yi |2

i∈I i∈I X X X X 2 2 + qi |xi | pi |yi | − 2 pi xi y¯i qi xi y¯i . i∈I

i∈I

i∈I

i∈I

By (CBS) −inequality, we have " # 21 X X X X X X pi xi y¯i qi xi y¯i ≤ pi |xi |2 pi |yi |2 qi |xi |2 qi |yi |2 i∈I

i∈I

i∈I

i∈I

i∈I

i∈I

and thus X i∈I

pi |xi |2

X i∈I

qi |yi |2 +

X

qi |xi |2

i∈I

X

pi |yi |2

i∈I

! 12

 ! 21 X X X − 2 pi xi y¯i qi xi y¯i ≥  pi |xi |2 i∈I

i∈I

X

qi |yi |2

i∈I

i∈I

! 12 2

! 12 −

X

qi |xi |2

i∈I

X

pi |yi |2

 . (3.23)

i∈I

Utilising (3.22) and (3.23) we deduce the desired inequality (3.21). The following corollary concerning a strong monotonicity result also holds [2, p. 18].

Corollary 113 For any p, q ∈ S+ (R) with p ≥ q one has the inequality: S¯ (p, I, x, y) − S¯ (q, I, x, y) 



kxkq,I

kykq,I

≥ det 

2  ≥ 0. (3.24)

kxkp−q,I

kykp−q,I

80

CHAPTER 3. FUNCTIONAL PROPERTIES

Remark 114 The following refinement of the (CBS) −inequality is a natural consequence of (3.21) [2, p. 19]

X i∈I

2 X X 2 2 |xi | |yi | − xi y¯i i∈I

i∈I

2 X |xi |2 sin2 αi |yi |2 sin2 αi − xi y¯i sin2 αi ≥ i∈I i∈I i∈I 2 X X X 2 2 2 2 2 |xi | cos αi |yi | cos αi − + xi y¯i cos αi i∈I i∈I i∈I      12 2  21 P 2 2 P 2 2 |yi | sin αi |xi | sin αi       i∈I i∈I    + det       12   12     P P 2 |xi |2 cos 2 αi |yi | cos2 αi X

X

i∈I

i∈I

≥ 0. (3.25) where αi ∈ R, i ∈ I.

3.5

Strong Superadditivity as an Index Set Mapping

We assume that we are under the hypothesis and notations in Section 3.2. Reconsider the functional S¯ (·, ·, ·, ·) : S+ (R) × Pf (N) × S (K) × S (K) → R, S¯ (p, I, x, y) :=

X i∈I

2 X X pi |xi |2 pi |yi |2 − pi xi y¯i . i∈I

(3.26)

i∈I

The following strong supperadditivity property as an index set mapping holds [2, p. 18].

3.5. STRONG SUPERADDITIVITY AS AN INDEX SET MAPPING 81 Theorem 115 For any p ∈ S+ (R) , I, J ∈ Pf (N) \ {∅} with I ∩ J = ∅ and x, y ∈ S (K) , we have S¯ (p, I ∪ J, x, y) − S¯ (p, I, x, y) − S¯ (p, J, x, y)   kxkp,I ≥ det  kxkp,J

kykp,I

2  ≥ 0. (3.27)

kykp,J

Proof. We have S¯ (p, I ∪ J, x, y) ! ! X X X X 2 2 2 2 = pi |xi | + pj |xj | pi |yi | + pj |yj | i∈I

j∈J

i∈I

(3.28)

j∈J

2 X X pj xj y¯j − pi xi y¯i + j∈I i∈I X X X X ≥ pi |xi |2 pi |yi |2 + pj |xj |2 pj |yj |2 i∈I

+

i∈I

X i∈I

j∈J

j∈J

X X X pj |xj |2 pi |yi |2 pi |xi |2 pj |yj |2 + j∈J

i∈I

j∈J

!2 X X pj xj y¯j pi xi y¯i + − j∈I i∈I X X pi |xi |2 pj |yj |2 = S¯ (p, I, x, y) + S¯ (p, J, x, y) + i∈I

j∈J

X X X X 2 2 + pi |yi | pj |xj | − 2 pi xi y¯i pj xj y¯j . i∈I

j∈J

i∈I

j∈I

By the (CBS) −inequality, we have " # 21 X X X X X X 2 2 2 2 pj |yj | pj |xj | pi |yi | pj xj y¯j ≤ pi |xi | pi xi y¯i i∈I

j∈I

i∈I

i∈I

j∈J

j∈J

82

CHAPTER 3. FUNCTIONAL PROPERTIES

and thus X

pi |xi |2

i∈I

X

pj |yj |2 +

j∈J

X

pi |yi |2

i∈I

X

pj |xj |2

j∈J

 ! 12 X X X − 2 pi xi y¯i pj xj y¯j ≥  pi |xi |2 i∈I

j∈I

! 12 X

pj |yj |2

j∈J

i∈I

! 12 2

! 12 −

X

X

pi |yi |2

i∈I

pj |xj |2

 . (3.29)

j∈J

If we use now (3.28) and (3.29), we may deduce the desired inequality (3.27). The following corollary concerning strong monotonicity also holds [2, p. 18]. Corollary 116 For any I, J ∈ Pf (N) \ {∅} with I ⊇ J one has the inequality 



S¯ (p, I, x, y)− S¯ (p, J, x, y) ≥ det 

kxkp,J

kykp,J

2  ≥ 0. (3.30)

kxkp,I\J

kykp,I\J

Remark 117 The following refinement of the (CBS) −inequality is a natural consequence of (3.27) [2, p. 19]. Suppose pi ≥ 0, i ∈ {1, . . . , 2n} and xi , yi ∈ K, i ∈ {1, . . . , 2n} . Then we have the inequality 2n X i=1

2

pi |xi |

2n X i=1

2 2n X pi |yi | − pi xi y¯i 2

i=1

n 2 n X X ≥ p2i |x2i |2 p2i |y2i |2 − p2i x2i y¯2i i=1 i=1 i=1 2 n n n X X X 2 2 + p2i−1 |x2i−1 | p2i−1 |y2i−1 | − p2i−1 x2i−1 y¯2i−1 n X

i=1

i=1

i=1

3.6. ANOTHER SUPERADDITIVITY PROPERTY    12 n P 2 p2i |x2i |     i=1   + det      12 n   P 2 p2i−1 |x2i−1 | 



p2i |y2i |2

 12

2

    ≥ 0. (3.31) n  12   P p2i−1 |y2i−1 |2 i=1

i=1

3.6

n P

83

i=1

Another Superadditivity Property

Let Pf (N) be the family of finite parts of the set of natural numbers, S (R) the linear space of real sequences and S+ (R) the family of nonnegative real sequences. Consider the mapping C : S+ (R) × Pf (N) × S (R) × S (R) → R !2 C (p, I, a, b) :=

X

pi a2i

i∈I

X

pi b2i −

X

i∈I

p i ai b i

.

(3.32)

i∈I

The following identity holds [3, p. 115]. Lemma 118 For any p, q ∈ S+ (R) one has C (p + q, I, a, b) = C (p, I, a, b) + C (q, I, a, b) +

X

pi qj (ai bj − aj bi )2 . (3.33)

(i,j)∈I×I

Proof. Using the well-known Lagrange’s identity, we have C (p, I, a, b) =

1 2

X

pi pj (ai bj − aj bi )2 .

(i,j)∈I×I

Thus C (p + q, I, a, b) 1 X = (pi + qi ) (pj + qj ) (ai bj − aj bi )2 2 (i,j)∈I×I

(3.34)

84

CHAPTER 3. FUNCTIONAL PROPERTIES =

1 2

X

pi pj (ai bj − aj bi )2 +

(i,j)∈I×I

+

1 2

X

1 2

X

qi qj (ai bj − aj bi )2

(i,j)∈I×I

1 X pj qi (ai bj − aj bi )2 2 (i,j)∈I×I X pi qj (ai bj − aj bi )2

pi qj (ai bj − aj bi )2 +

(i,j)∈I×I

= C (p, I, a, b) + C (q, I, a, b) +

(i,j)∈I×I

since, by symmetry, X pi qj (ai bj − aj bi )2 = (i,j)∈I×I

X

pj qi (ai bj − aj bi )2 .

(i,j)∈I×I

Consider the following mapping:  X X 1 D (p, I, a, b) := [C (p, I, a, b)] 2 =  pi a2i pi b2i − i∈I

i∈I

!2  12 X

pi ai bi

 .

i∈I

The following result has been obtained in [4, p. 88] as a particular case of a more general result holding in inner product spaces. Theorem 119 For any p, q ∈ S+ (R) , I ∈ Pf (N) and a, b ∈ S (R) , we have the superadditive property D (p + q, I, a, b) ≥ D (p, I, a, b) + D (q, I, a, b) ≥ 0.

(3.35)

Proof. We will give here an elementary proof following the one in [3, p. 116 – p. 117]. By Lemma 118, we obviously have D2 (p + q, I, a, b) = D2 (p, I, a, b) + D2 (q, I, a, b) +

X

pi qj (ai bj − aj bi )2 . (3.36)

(i,j)∈I×I

We claim that X (i,j)∈I×I

pi qj (ai bj − aj bi )2 ≥ 2D (p, I, a, b) D (q, I, a, b) .

(3.37)

3.6. ANOTHER SUPERADDITIVITY PROPERTY

85

Taking the square in both sides of (3.37), we must prove that #2

" X

pi a2i

X

qi b2i +

X i∈I

i∈I

i∈I

qi a2i

X

pi b2i −2

i∈I

X i∈I

 X X pi a2i pi b2i − ≥ 4 i∈I

p i ai b i

X

q i ai b i

i∈I

!2  X

i∈I

p i ai b i



i∈I

 X X × qi a2i qi b2i − i∈I

!2  X

i∈I

q i ai b i

 . (3.38)

i∈I

Let us denote ! 21 a :=

X

pi a2i

! 12 ,

x :=

X

i∈I

qi a2i

! 21 ,

b :=

i∈I

X

pi b2i

,

i∈I

! 21 y :=

X i∈I

qi b2i

,

c :=

X

p i ai b i ,

z :=

i∈I

X

q i ai b i .

i∈I

With these notations (3.38) may be written in the following form
2

a2 y 2 + b2 x2 − 2cz ≥ 4 a2 b2 − c2 x2 y 2 − z 2 .

(3.39)

Using the elementary inequality

m2 − n2 p2 − q 2 ≤ (mp − nq)2 , m, n, p, q ∈ R we may state that 4 (abxy − cz)2 ≥ 4 a2 b2 − c2





x2 y 2 − z 2 ≥ 0.

(3.40)

Since, by the (CBS) −inequality, we observe that abxy ≥ |cz| ≥ |cz| , we can state that a2 y 2 + b2 x2 − 2cz ≥ 2 (abxy − cz) ≥ 0 giving a2 y 2 + b2 x2 − 2cz


2

≥ 4 (abxy − cz)2 .

(3.41)

Utilizing (3.40) and (3.41) we deduce the inequality (3.39), and (3.37) is proved.

86

CHAPTER 3. FUNCTIONAL PROPERTIES Finally, by (3.36) and (3.37) we have D2 (p + q, I, a, b) ≥ [D (p, I, a, b) + D (q, I, a, b)]2 ,

i.e., the superadditivity property (3.35). Remark 120 The following refinement of the (CBS) − inequality holds [4, p. 89]  X X  a2i b2i − i∈I

i∈I

!2  12 X

ai bi



i∈I

 X X a2i sin2 αi ≥ b2i sin2 αi − i∈I

i∈I

X

ai bi sin2 αi



i∈I

 X X + a2i cos 2 αi b2i cos2 αi − i∈I

!2  12

i∈I

!2  12 X

ai bi cos2 αi



i∈I

≥ 0 (3.42) for any αi ∈ R, i ∈ {1, . . . , n} .

3.7

The Case of Index Set Mapping

Assume that we are under the hypothesis and notations in Section 3.6. Reconsider the functional C : S+ (R) × Pf (N) × S (R) × S (R) → R given by !2 X X X 2 2 C (p, I, a, b) := p i ai p i bi − p i ai b i . (3.43) i∈I

i∈I

i∈I

The following identity holds. Lemma 121 For any I, J ∈ Pf (N) \ {∅} with I ∩J 6= ∅ one has the identity: C (p, I ∪ J, a, b) = C (p, I, a, b) + C (p, J, a, b) +

X (i,j)∈I×J

pi pj (ai bj − aj bi )2 . (3.44)

3.7. THE CASE OF INDEX SET MAPPING

87

Proof. Using Lagrange’s identity [5, p. 84], we may state 1 X C (p, K, a, b) = pi pj (ai bj − aj bi )2 , K ∈ Pf (N) \ {∅} . 2

(3.45)

(i,j)∈K×K

Thus C (p, I ∪ J, a, b) X 1 = pi pj (ai bj − aj bi )2 2 (i,j)∈(I∪J)×(I∪J)

=

1 2

X

pi pj (ai bj − aj bi )2 +

(i,j)∈I×I

1 + 2

X

1 2

X

pi pj (ai bj − aj bi )2

(i,j)∈I×J

1 X pi pj (ai bj − aj bi )2 2 (i,j)∈J×J X pi pj (ai bj − aj bi )2

pi pj (ai bj − aj bi )2 +

(i,j)∈J×I

= C (p, I, a, b) + C (p, J, a, b) +

(i,j)∈I×J

since, by symmetry, X pi pj (ai bj − aj bi )2 = (i,j)∈I×J

X

pi pj (ai bj − aj bi )2 .

(i,j)∈J×I

Now, if we consider the mapping  X X 1 D (p, I, a, b) := [C (p, I, a, b)] 2 =  pi a2i pi b2i − i∈I

i∈I

!2  21 X

p i ai b i

 ,

i∈I

then the following superadditivity property as an index set mapping holds: Theorem 122 For any I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅ one has D (p, I ∪ J, a, b) ≥ D (p, I, a, b) + D (p, J, a, b) ≥ 0.

(3.46)

Proof. By Lemma 121, we have D2 (p, I ∪ J, a, b) = D2 (p, I, a, b) + D2 (p, J, a, b) +

X (i,j)∈I×J

pi pj (ai bj − aj bi )2 (3.47)

88

CHAPTER 3. FUNCTIONAL PROPERTIES

To prove (3.46) it is sufficient to show that X

pi pj (ai bj − aj bi )2 ≥ 2D (p, I, a, b) D (p, J, a, b) .

(3.48)

(i,j)∈I×J

Taking the square in (3.48), we must demonstrate that #2

" X i∈I

pi a2i

X

pj b2j +

X

pj a2j

j∈J

j∈J

X

pi b2i − 2

X

 X X pi a2i pi b2i − ≥ 4 i∈I

p i ai b i

i∈I

i∈I

i∈I

X

p j aj b j

j∈J

!2  X

p i ai b i



i∈I

 X X × pj a2j pj b2j − j∈J

j∈J

!2  X

p j aj b j

.

j∈J

If we denote ! 21 a :=

X

pi a2i

! 12 ,

X

x :=

i∈I

pj a2j

! 12 ,

b :=

j∈J

X

pi b2i

,

i∈I

! 21 y :=

X

pj b2j

,

c :=

j∈J

X

p i ai b i ,

z :=

i∈I

X

p j aj b j ,

j∈J

then we need to prove a2 y 2 + b2 x2 − 2cz


2

≥ 4 a2 b 2 − c 2





x2 y 2 − z 2 ,

(3.49)

which has been shown in Section 3.6. This completes the proof. Remark 123 The following refinement of the (CBS) −inequality holds  2n 2n X X 2  p i ai pi b2i − i=1

i=1

2n X i=1

!2  12 p i ai b i



3.8. SUPERMULTIPLICITY IN TERMS OF WEIGHTS  n n X X 2  ≥ p2i a2i p2i b22i − i=1

n X

i=1

3.8

!2  12 p2i a2i b2i 

i=1

 n n X X + p2i−1 a22i−1 p2i−1 b22i−1 − i=1

89

i=1

n X

!2  21 p2i−1 a2i−1 b2i−1

 ≥ 0.

i=1

Supermultiplicity in Terms of Weights

Denote by S+ (R) the set of nonnegative sequences. Assume that A : S+ (R) → R is additive on S+ (R) , i.e., A (p + q) = A (p) + A (q) ,

p, q ∈ S+ (R)

(3.50)

and L : S+ (R) → R is superadditive on S+ (R) , i.e., L (p + q) ≥ L (p) + L (q) ,

p, q ∈ S+ (R) .

(3.51)

L (p) and H (p) := [F (p)]A(p) . A (p)

(3.52)

Define the following associated functionals F (p) :=

The following result holds [3, Theorem 2.1]. Lemma 124 With the above assumptions, we have H (p + q) ≥ H (p) H (q) ;

(3.53)

for any p, q ∈ S+ (R) , i.e., H (·) is supermultiplicative on S+ (R) . Proof. We shall follow the proof in [3]. Using the well-known arithmetic mean-geometric mean inequality for real numbers β α αx + βy (3.54) ≥ x α+β y α+β α+β

90

CHAPTER 3. FUNCTIONAL PROPERTIES

for any x, y ≥ 0 and α, β ≥ 0 with α + β > 0, we have successively L (p + q) L (p + q) L (p) + L (q) F (p + q) = = ≥ A (p + q) A (p) + A (q) A (p) + A (q) =

L(p) L(q) A (p) A(p) + A (q) A(q)

A (p) + A (q)

A (p) F (p) + A (q) F (q) A (p) + A (q)

=

A(p)

(3.55)

A(q)

≥ [F (p)] A(p)+A(q) · [F (q)] A(p)+A(q) for all p, q ∈ S+ (R) . However, A (p) + A (q) = A (p + q) , and thus (3.55) implies the desired inequality (3.53). We are now able to point out the following inequality related to the (CBS) − inequality. The first result is incorporated in the following theorem [3, p. 115]. Theorem 125 For any p, q ∈ S+ (R) , and a, b ∈ S (R) , one has the inequality ( " X X 1 (pi + qi ) a2i (pi + qi ) b2i PI + QI i∈I i∈I !2 PI +QI  X  − (pi + qi ) ai bi  i∈I   !2 PI 1 X  X X  ≥ pi a2i pi b2i − p i ai b i    PI i∈I i∈I i∈I   !2 QI   1 X X X  × qi a2i qi b2i − q i ai b i   QI  i∈I

i∈I

i∈I

> 0, (3.56) where PI :=

P

i∈I

pi > 0, QI :=

P

i∈I

qi > 0.

Proof. Consider the functionals X A (p) := p i = PI ; i∈I

!2 C (p) :=

X i∈I

pi a2i

X i∈I

pi b2i −

X i∈I

p i ai b i

.

3.8. SUPERMULTIPLICITY IN TERMS OF WEIGHTS

91

Then A (·) is additive and C (·) is superadditive (see for example Lemma 118) on S+ (R) . Applying Lemma 124 we deduce the desired inequality (3.56). The following refinement of the (CBS) −inequality holds. Corollary 126 For any a, b, α ∈ S (R) , one has the inequality !2 n n n X X X a2i b2i − ai b i i=1

i=1

i=1

1

≥ 1 n

Pn

2 i=1 sin αi


n1 Pni=1 sin2 αi

 n n X X 2 2  × ai sin αi b2i sin2 αi − i=1

i=1


1 Pn 2 2 α n i=1 cos αi cos i i=1 P !2  n1 ni=1 sin2 αi

Pn

n X

ai bi sin2 αi



i=1

 n n X X 2 2  × ai cos αi b2i cos2 αi − i=1

1 n

n X

!2  n1 ai bi cos2 αi

Pn

i=1

cos2 αi



i=1

i=1

≥ 0. (3.57) The following result holds [3, p. 116]. Theorem 127 For any p, q ∈ S+ (R) , and a, b ∈ S (R) , one has the inequality " # 21 " # 12  X X 1 1 (pi + qi ) a2i (pi + qi ) b2i  PI + QI P + Q I I i∈I i∈I )PI +QI X 1 − (pi + qi ) ai bi PI + QI i∈I  PI ! 12 ! 21 1 X 2 1 X 2 1 X ≥ p i ai p i bi − p i ai b i  PI PI i∈I PI i∈I i∈I  QI ! 12 ! 12 X X X 1 1 1 2 2  × q i ai qi b i − q i ai b i  QI QI QI i∈I

i∈I

i∈I

≥ 0. (3.58)

92

CHAPTER 3. FUNCTIONAL PROPERTIES Proof. Follows by Lemma 124 on taking into account that the functional ! 12 ! 12 X X X 2 2 p i bi − p i ai b i B (p) := p i ai i∈I

i∈I

i∈I

is superadditive on S+ (R) (see Section 3.2). The following refinement of the (CBS) −inequality holds. Corollary 128 For any a, b, α ∈ S (R) , one has the inequality n X

! 12 a2i

i=1

b2i

i=1

n X

Pn

2 i=1 sin αi ! 12

a2i sin2 αi

×

n X


n1 Pni=1 sin2 αi ·

1


1 Pn 2 2 α n i=1 cos αi cos i i=1 P  n1 ni=1 sin2 αi ! 12 n n X X b2i sin2 αi − ai bi sin2 αi  1 n

i=1

i=1



n X − ai b i 1

1 n

×

! 12

i=1

≥ 

n X

! 12 a2i cos2 αi

i=1

n X

Pn

i=1

! 12 b2i cos2 αi

i=1

 n1 n X − ai bi cos2 αi 

Pn

i=1

cos2 αi

i=1

≥ 0. (3.59) Finally, we may also state [3, p. 117]. Theorem 129 For any p, q ∈ S+ (R) , and a, b ∈ S (R) , one has the inequality " X X 1 1 (pi + qi ) a2i · (pi + qi ) b2i PI + QI i∈I PI + QI i∈I I !2  PI +Q 2 X 1 − (pi + qi ) ai bi  PI + QI i∈I  ≥

1 X 2 1 X 2 p i ai · p i bi − PI i∈I PI i∈I

1 X p i ai b i PI i∈I

!2  P2I 

3.9. SUPERMULTIPLICITY AS AN INDEX SET MAPPING  ×

1 X 2 1 X 2 q i ai · qi b i − QI i∈I QI i∈I

1 X q i ai b i QI i∈I

93

!2  Q2I . (3.60)



Proof. Follows by Lemma 124 on taking into account that the functional  n n X X 2 pi b2i − D (p) :=  p i ai

n X

p i ai b i



i=1

i=1

i=1

!2  21

is superadditive on S+ (R) (see Section 3.6). The following corollary also holds. Corollary 130 For any a, b, α ∈ S (R) , one has the inequality  n n X X 2  b2i − ai i=1

n X

!2  12 ai b i

i=1

i=1

1

≥ 1 n



Pn

2


n1 Pni=1 sin2 αi ·

i=1 sin αi  n n X X 2 2  × ai sin αi b2i sin2 αi − i=1

i=1

 n n X X × a2i cos2 αi b2i cos2 αi − i=1

i=1

1 1 n

Pn

n X

i=1

cos2


n1 Pni=1 cos 2 αi

αi P !2  2n1 ni=1 sin2 αi

ai bi sin2 αi



i=1 n X

!2  2n1 ai bi cos2 αi 

Pn

i=1

cos2 αi

i=1

≥ 0. (3.61)

3.9

Supermultiplicity as an Index Set Mapping

Denote by Pf (N) the set of all finite parts of the natural number set N and assume that B : Pf (N) → R is set-additive on Pf (N) , i.e., B (I ∪ J) = B (I) + B (J) for any I, J ∈ Pf (N) , I ∩ J 6= ∅,

(3.62)

94

CHAPTER 3. FUNCTIONAL PROPERTIES

and G : Pf (N) → R is set-superadditive on Pf (N) , i.e., G (I ∪ J) ≥ G (I) + G (J) for any I, J ∈ Pf (N) , I ∩ J 6= ∅.

(3.63)

We may define the following associated functionals M (I) :=

G (I) and N (I) := [M (I)]A(I) . B (I)

(3.64)

With these notations we may prove the following lemma that is interesting in itself as well. Lemma 131 Under the above assumptions one has N (I ∪ J) ≥ N (I) N (J)

(3.65)

for any I, J ∈ Pf (N) \ {∅} with I ∩J 6= ∅, i.e., N (·) is set-supermultiplicative on Pf (N) . Proof. Using the arithmetic mean – geometric mean inequality β α αx + βy ≥ x α+β y α+β α+β

(3.66)

for any x, y ≥ 0 and α, β ≥ 0 with α + β > 0, we have successively for I, J ∈ Pf (N) \ {∅} with I ∩ J 6= ∅ that M (I ∪ J) = =

G (I ∪ J) G (I ∪ J) G (I) + G (J) = ≥ B (I ∪ J) B (I) + B (J) B (I) + B (J) G(I) G(J) B (I) B(I) + B (J) B(J)

B (I) + B (J) B(I)

=

(3.67)

B (I) M (I) + B (J) M (J) B (I) + B (J) B(J)

≥ (M (I)) B(I)+B(J) · (M (J)) B(I)+B(J) . Since B (I) + B (J) = B (I ∪ J) , we deduce by (3.67) the desired inequality (3.65). Now, we are able to point out some set-superadditivity properties for some functionals associates to the (CBS) −inequality. The first result is embodied in the following theorem.

3.9. SUPERMULTIPLICITY AS AN INDEX SET MAPPING

95

Theorem 132 If a, b ∈ S (R) , p ∈ S+ (R) and I, J ∈ Pf (N) \ {∅} so that I ∩ J 6= ∅, then one has the inequality   !2 PI∪J   1 X X X 2 2   p k bk − p k ak b k p k ak   PI∪J k∈I∪J k∈I∪J k∈I∪J   !2 PI 1 X  X X 2 2   ≥ p i ai p i bi − p i ai b i  PI  i∈I i∈I i∈I   !2 PJ   1 X X X  × pj a2j pj b2j − pj aj bj  , (3.68)  PJ  j∈J

when PJ :=

P

j∈J

j∈J

j∈J

pj .

Proof. Consider the functionals X B (I) := pi ; i∈I

!2 G (I) :=

X i∈I

pi a2i

X i∈I

pi b2i −

X

p i ai b i

.

i∈I

The functional B (·) is obviously set-additive and (see Section 3.7) the functional G (·) is set-superadditive. Applying Lemma 131 we then deduce the desired inequality (3.68). The following corollary is a natural application. Corollary 133 If a, b ∈ S (R) and p ∈ S+ (R) , then for any n ≥ 1 one has the inequality   !2 P2n 2n 2n 2n  1 X  X X  pi a2i pi b2i − p i ai b i   P2n  i=1 i=1 i=1 P   !2  ni=1 p2i n n n   X X X 1  ≥ Pn p2i a22i p2i b22i − p2i a2i b2i   i=1 p2i  i=1 i=1 i=1

96

CHAPTER 3. FUNCTIONAL PROPERTIES ( ×

" n n X X 1 2 Pn p2i−1 a2i−1 p2i−1 b22i−1 i=1 p2i−1 i=1 i=1 Pni=1 p2i−1 ! 2 n  X − p2i−1 a2i−1 b2i−1  . (3.69)  i=1

The following result also holds. Theorem 134 If a, b ∈ S (R) , p ∈ S+ (R) and I, J ∈ Pf (N) \ {∅} so that I ∩ J 6= ∅, then one has the inequality " # 12 " # 12  1 X X 1 pk a2k pk b2k  PI∪J P I∪J k∈I∪J k∈I∪J )PI∪J 1 X − p k ak b k PI∪J k∈I∪J  PI ! 12 ! 12  1 X  1 X 2 1 X 2 p i ai p i bi p i ai b i ≥ −  PI PI  PI i∈I i∈I i∈I  PJ ! 12 ! 12   1 X X X 1 1 × pj a2j pj b2j − p j aj b j . (3.70)  PJ PJ  PJ j∈J j∈J j∈J Proof. Follows by Lemma 131 on taking into account that the functional ! 12

! 12 G (I) :=

X i∈I

pi a2i

X i∈I

pi b2i

X − p i ai b i i∈I

is set-superadditive on Pf (N) . The following corollary is a natural application. Corollary 135 If a, b ∈ S (R) and p ∈ S+ (R) , then for any n ≥ 1 one has

3.9. SUPERMULTIPLICITY AS AN INDEX SET MAPPING the inequality  ! 12 2n X 1  pi a2i P2n i=1 

2n 1 X 2 p i bi P2n i=1

1

n X

! 12

97

P2n 2n 1 X − p i ai b i  P2n i=1

!

1 2

1

p2i b22i p p 2i 2i i=1 i=1 i=1 i=1 Pn # n i=1 p2i X P 1 − n p2i a2i b2i i=1 p2i i=1  ! 12 ! 12 n n X X 1 1 Pn ×  Pn p2i−1 a22i−1 p2i−1 b22i−1 p p i=1 2i−1 i=1 i=1 2i−1 i=1 #Pni=1 p2i−1 n X P 1 − n p2i−1 a2i−1 b2i−1 . (3.71) i=1 p2i−1 ≥  Pn

p2i a22i

! 12

n X

Pn

i=1

Finally, we may also state: Theorem 136 If a, b ∈ S (R) , p ∈ S+ (R) and I, J ∈ Pf (N) \ {∅} so that I ∩ J 6= ∅, then one has the inequality !2  PI∪J 2

 X 1 X  1 pk a2k · pk b2k − PI∪J k∈I∪J PI∪J k∈I∪J

1 PI∪J

X

p k ak b k

k∈I∪J

 ≥

1 X 2 1 X 2 p i ai · pi bi − PI i∈I PI i∈I



1 X p i ai b i PI i∈I

!2  P2I 

 ×

1 X 1 X 2 pj a2j · pj bj − PJ j∈J PJ j∈J

1 X p j aj b j PJ j∈J

!2  P2J 

. (3.72)

Proof. Follows by Lemma 131 on taking into account that the functional  !2  12 X X X Q (I) :=  pi a2i pi b2i − p i ai b i  i∈I

i∈I

i∈I

98

CHAPTER 3. FUNCTIONAL PROPERTIES

is set-superadditive on Pf (N) (see Section 3.7). The following corollary holds as well. Corollary 137 If a, b ∈ S (R) and p ∈ S+ (R) , then for any n ≥ 1 one has the inequality   1 P2n

2n X

1 P2n "

pi a2i ·

i=1

≥

2n X

pi b2i −

i=1 n X

1

Pn

i=1 p2i

Pn

i=1

" ×

1 Pn i=1 p2i−1 −

n X i=1

1 P2n

!2  P2n 2 p i ai b i

p2i a22i n X

p2i



i=1

!

!

n X

1

p2i b22i p i=1 2i i=1 P !  12 ni=1 p2i

Pn

i=1

1

−

2n X

2

p2i a2i b2i



i=1

! p2i−1 a22i−1

! n X 1 Pn p2i−1 b22i−1 p i=1 2i−1 i=1 P !  12 ni=1 p2i−1

n X 1 Pn p2i−1 a2i−1 b2i−1 i=1 p2i−1 i=1

2



. (3.73)

Bibliography ˇ ARSLANAGIC, ´ The improvement of Cauchy[1] S.S. DRAGOMIR and S.Z. Buniakowski-Schwartz’s inequality, Mat. Bilten, 16 (1992), 77-80. [2] S.S. DRAGOMIR and B. MOND, On the superadditivity and monotonicity of Schwartz’s inequality in inner product spaces, Contributions, Ser. Math. Sci. Macedonian Acad. Sci., 15(1) (1994), 5-22. [3] R.P. AGARWAL and S.S. DRAGOMIR, The property of supermultiplicity for some classical inequalities and applications, Computer Math. Applic., 35(6) (1998), 105-118. [4] S.S. DRAGOMIR and B. MOND, On the superadditivity and monotonicity of Gram’s inequality and related results, Acta Math. Hungarica, 71(1-2) (1996), 75-90. ´ J.E. PECARI ˘ ´ and A.M. FINK, Classical and [5] D.S. MITRINOVIC, C New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.

99

100

BIBLIOGRAPHY

Chapter 4 Counterpart Inequalities 4.1

The Cassels’ Inequality

The following result was proved by J.W.S. Cassels in 1951 (see Appendix 1 of [2] or Appendix of [3]): ¯ = (b1 , . . . , bn ) be sequences of positive Theorem 138 Let ¯ a = (a1 , . . . , an ) , b real numbers and w ¯ = (w1 , . . . , wn ) a sequence of nonnegative real numbers. Suppose that     ai ai and M = max . (4.1) m = min bi bi i=1,n i=1,n Then one has the inequality Pn

Pn 2 2 w a (m + M )2 i i i=1 i=1 wi bi ≤ . P 2 4mM ( ni=1 wi ai bi )

The equality holds in (4.2) when w1 = m = abn1 and M = ban1 .

1 , a1 b1

wn =

1 , an bn

(4.2)

w2 = · · · = wn−1 = 0,

Proof. 1. The original proof by Cassels (1951) is of interest. We shall follow the paper [5] in sketching this proof. We begin with the assertion that (1 + kw) (1 + k −1 w) (1 + k) (1 + k −1 ) ≤ , k > 0, w ≥ 0 4 (1 + w)2 101

(4.3)

102

CHAPTER 4. COUNTERPART INEQUALITIES

which, being an equivalent form of (4.2) for n = 2, shows that it holds for n = 2. To prove that the maximum of (4.2) is obtained when we have more than two wi ’s being nonzero, Cassels then notes that if for example, w1 , w2 , w3 6= 0 , then we would have the linear equations lead to an extremum M of XY 22 a2n X + b2n Y − 2M an bn Z = 0, k = 1, 2, 3. Nontrivial solutions exist if and only if the three vectors [a2n , b2n , an bn ] are linearly dependent. But this will be so only if, for some i 6= j (i, j = 1, 2, 3) ai = γaj , bi = γbj . And if that were true, we could, for example, drop the ai , bi terms and so deal with the same problem with one less variable. If only one wi 6= 0, then M = 1, the lower bound. So we need only examine all pairs wi 6= 0, wj 6= 0. The result (4.2) then quickly follows. 2. We will now use the barycentric method of Frucht [1] and Watson [4]. We will follow the paper [5]. We substitute wi = ub2i in the left hand side of (4.2), which may then be i expressed as the ratio N D2 where n  2 n   X X ai ai N= ui and D = ui , b b i i i=1 i=1 Pn assuming without loss of generality, that i=1 ai = 1. But the pointwith co 2 ai ai ordinates (D, N ) must lie within the convex closure of the n points bi , b2 . n oi N The value of D2 at points on the parabola is one unit. If m = min abii and i=1,n n o M = max abii , then the minimum must lie on the chord joining the point i=1,n

(m, m2 ) and (M, M 2 ) . Some easy calculus then leads to (4.2). The following “unweighted” Cassels’ inequality holds. ¯ satisfy the assumptions in Theorem 138, one has Corollary 139 If ¯ a and b the inequality Pn 2 Pn 2 (m + M )2 i=1 ai i=1 bi ≤ . (4.4) P 2 4mM ( ni=1 ai bi ) The following two additive versions of Cassels inequality hold.

4.1. THE CASSELS’ INEQUALITY

103

Corollary 140 With the assumptions of Theorem 138, one has n X

0≤

wi a2i

i=1

√ ≤

n X

! 12 wi b2i

−

n X

wi ai bi

(4.5)

i=1

i=1

√ 2 n M− m X √ wi ai bi . 2 mM i=1

and 0≤

n X

wi a2i

i=1

n X

wi b2i

−

n X

i=1 n X

(M − m)2 ≤ 4mM

!2 wi ai bi

(4.6)

i=1 !2

wi ai bi

.

i=1

Proof. Taking the square root in (4.2) we get 1 P P ( ni=1 wi a2i ni=1 wi b2i ) 2 M +m Pn 1≤ ≤ √ . 2 mM i=1 wi ai bi

Subtracting 1 on both sides, a simple calculation will lead to (4.5). The second inequality follows by (4.2) on subtracting 1 and appropriate computation. The following additive version of unweighted Cassels inequality also holds. ¯ one has Corollary 141 With the assumption of Theorem 138 for ¯ a and b the inequalities

0≤

n X i=1

a2i

n X

! 21 b2i

−

n X

i=1

√ ai b i ≤

i=1

√ 2 n M− m X √ ai b i 2 mM i=1

(4.7)

and 0≤

n X i=1

a2i

n X i=1

b2i −

n X i=1

!2 ai b i

(M − m)2 ≤ 4mM

n X i=1

!2 ai b i

.

(4.8)

104

4.2

CHAPTER 4. COUNTERPART INEQUALITIES

The P´ olya-Szeg¨ o Inequality

The following inequality was proved in 1925 by P´olya and Szeg¨o [6, pp. 57, 213 – 214], [7, pp. 71– 72, 253 – 255]. ¯ = (b1 , . . . , bn ) be two sequences of Theorem 142 Let ¯ a = (a1 , . . . , an ) and b positive real numbers. If 0 < a ≤ ai ≤ A < ∞, 0 < b ≤ bi ≤ B < ∞ for each i ∈ {1, . . . , n} , (4.9) then one has the inequality Pn

a2i

Pn

2 i=1 bi Pn 2 ( i=1 ai bi ) i=1

(ab + AB)2 ≤ . 4abAB

(4.10)

The equality holds in (4.10) if and only if     A B B A B A + and q = n · + p=n· a a b b a b are integers and if p of the numbers a1 , . . . , an are equal to a and q of these numbers are equal to A, and if the corresponding numbers bi are equal to B and b respectively. Proof. Following [5], we shall present here the original proof of P´olya and Szeg¨o. We may, without loss of generality, suppose that a1 ≥ · · · ≥ an , then to maximise the left-hand side of (4.10) we must have that the critical bi ’s be reversely ordered (for if bk > bm with k < m, then we can interchange bk and bm such that b2k + b2m = b2m + b2k and ak bk + am bm ≥ ak bm + am bk ), i.e., that b1 ≤ · · · ≤ bn . P´olya and Szeg¨o then continue by defining nonnegative numbers ui and vi for i = 1, . . . , n − 1 and n > 2 such that a2i = ui a21 + vi a2n and b2i = ui b21 + vi b2n . Since ai bi > ui a1 b1 + vi an bn the left hand side of (4.10), Pn 2 Pn 2 (U a21 + V a2n ) (U b21 + V b2n ) i=1 bi i=1 ai ≤ , Pn 2 (U a1 b1 + V an bn )2 ( i=1 ai bi )

(4.11)

´ ¨ INEQUALITY 4.2. THE POLYA-SZEG O

105

P P where U = ni=1 ui and V = ni=1 vi . This reduces the problem to that with n = 2, which is solvable by elementary methods, leading to Pn 2 Pn 2 (a1 b1 + an bn )2 i=1 ai i=1 bi ≤ , (4.12) P 2 4a1 an b1 bn ( ni=1 ai bi ) where, since the ai ’s and bi ’s here are reversely ordered, a1 = max {ai } ,

an = min {ai } ,

b1 = min {bi } ,

i=1,n

i=1,n

i=1,n

bn = max {bi } . (4.13) i=1,n

0 < a ≤ ai ≤ A, 0 < b ≤ bi ≤ B,

i = (1, . . . , n) ,

If we now assume, as in (4.9), that

then

(ab + AB)2 (a1 b1 + an bn )2 ≤ 4a1 an b1 bn 4abAB

(because

(k+1)2 4k

≤

(α+1)2 4α

for k ≤ α), and the inequality (4.10) is proved.

Remark 143 The inequality (4.10) may also be obtained from the “unweighted” Cassels’ inequality Pn 2 Pn 2 (m + M )2 i=1 ai i=1 bi ≤ , (4.14) Pn 2 4mM ( i=1 ai bi ) where 0 < m ≤

ai bi

≤ M for each i ∈ {1, . . . , n} .

The following additive versions of the P´olya-Szeg¨o inequality also hold. Corollary 144 With the assumptions in Theorem 142, one has the inequality √ √ 2 ! 12 n n n n AB − ab X X X X 2 2 √ − ai b i ≤ ai bi (4.15) 0≤ ai bi 2 abAB i=1 i=1 i=1 i=1 and 0≤

n X i=1

a2i

n X i=1

b2i −

n X i=1

!2 ai b i

(AB − ab)2 ≤ 4abAB

n X i=1

!2 ai b i

.

(4.16)

106

4.3

CHAPTER 4. COUNTERPART INEQUALITIES

The Greub-Rheinboldt Inequality

The following weighted version of the P´olya-Szeg¨o inequality was obtained by Greub and Rheinboldt in 1959, [6]. ¯ = (b1 , . . . , bn ) be two sequences of Theorem 145 Let ¯ a = (a1 , . . . , an ) and b positive real numbers and w ¯ = (w1 , . . . , wn ) a sequence of nonnegative real numbers. Suppose that 0 < a ≤ ai ≤ A < ∞, 0 < b ≤ bi ≤ B < ∞

(i = 1, . . . , n) .

Then one has the inequality Pn Pn 2 2 (ab + AB)2 i=1 wi ai i=1 wi bi ≤ . P 2 4abAB ( ni=1 wi ai bi )

(4.17)

(4.18)

Equality holds in (4.18) when wi = a11b1 , wn = an1bn , w2 = · · · = wn−1 = 0, m = abn1 , M = ban1 with a1 = A, an = a, b1 = b and bn = b. Remark 146 This inequality follows by Cassels’ result which states that Pn Pn 2 2 w a (m + M )2 i i i=1 wi bi i=1 ≤ , (4.19) P 2 4mM ( ni=1 wi ai bi ) provided 0 < m ≤

ai bi

≤ M < ∞ for each i ∈ {1, . . . , n} .

The following additive versions of Greub-Rheinboldt also hold. Corollary 147 With the assumptions in Theorem 145, one has the inequalities ! 12 n n n X X X 2 2 0≤ wi ai wi bi − wi ai bi (4.20) i=1

√ ≤

i=1

i=1

√ 2 n AB − ab X √ wi ai bi 2 abAB i=1

and 0≤

n X i=1

wi a2i

n X

wi b2i −

i=1

(AB − ab)2 ≤ 4abAB

n X

!2 wi ai bi

i=1 n X i=1

!2 wi ai bi

.

(4.21)

4.4. A CASSELS’ TYPE INEQUALITY FOR COMPLEX NUMBERS 107

4.4

A Cassels’ Type Inequality for Complex Numbers

The following counterpart inequality for the (CBS) −inequality holds [9]. Theorem 148 Let a, A ∈ K (K = C, R) such that Re (¯ aA) > 0. If x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) are sequences of complex numbers and w ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers with the property that n X wi Re [(Ayi − xi ) (¯ xi − a ¯y¯i )] ≥ 0, (4.22) i=1

then one has the inequality "

n X

n X 2 wi |xi | wi |yi |2

i=1

i=1

The constant one.

1 2

# 21

1 ≤ · 2

Pn

i=1

wi Re [A¯ xi yi + a ¯xi y¯i ] 1

is sharp in the sense that it cannot be replaced by a smaller

Proof. We have, obviously, that Γ :=

n X

wi Re [(Ayi − xi ) (¯ xi − a ¯y¯i )]

i=1

=

n X

wi Re [A¯ xi yi + a ¯xi y¯i ] −

i=1

n X

2

wi |xi | − Re (¯ aA)

i=1

n X

wi |yi |2

i=1

and then, by (4.22), one has n X i=1

(4.23)

[Re (¯ aA)] 2 n X 1 |A| + |a| wi xi y¯i . ≤ · 1 2 [Re (¯ aA)] 2 i=1

2

wi |xi | + Re (¯ aA)

n X i=1

2

wi |yi | ≤

n X i=1

wi Re [A¯ xi yi + a ¯xi y¯i ]

108

CHAPTER 4. COUNTERPART INEQUALITIES

giving n X

1 [Re (¯ aA)]

1 2

2

wi |xi | + [Re (¯ aA)]

i=1

1 2

n X

wi |yi |2

i=1

Pn

i=1

≤

wi Re [A¯ xi yi + a ¯xi y¯i ] 1

. (4.24)

[Re (¯ aA)] 2 On the other hand, by the elementary inequality 1 2 q ≥ 2pq α

αp2 +

holding for any p, q ≥ 0 and α > 0, we deduce

2

n X

n X 2 wi |xi | wi |yi |2

i=1

i=1

! 12

n X

1

≤

[Re (¯ aA)]

1 2

2

wi |xi | + [Re (¯ aA)]

i=1

1 2

n X

wi |yi |2 . (4.25)

i=1

Utilising (4.24) and (4.25), we deduce the first part of (4.23). The second part is obvious by the fact that for z ∈ C, |Re (z)| ≤ |z| . Now, assume that the first inequality in (4.23) holds with a constant c > 0, i.e., n X

2

wi |xi |

i=1

n X

2

wi |yi | ≤ c ·

i=1

Pn

i=1

wi Re [A¯ xi yi + a ¯xi y¯i ] 1

,

(4.26)

[Re (¯ aA)] 2

where a, A, x ¯, y ¯ satisfy (4.22). If we choose a = A = 1, y = x 6= 0, then obviously (4.23) holds and from (4.26) we may get n n X X 2 wi |xi | ≤ 2c wi |xi |2 , i=1 1 . 2

i=1

giving c ≥ The theorem is completely proved. The following corollary is a natural consequence of the above theorem.

4.4. A CASSELS’ TYPE INEQUALITY FOR COMPLEX NUMBERS 109 Corollary 149 Let m, M > 0 and x ¯, y ¯, w ¯ be as in Theorem 148 and with the property that n X

wi Re [(M yi − xi ) (¯ xi − m¯ yi )] ≥ 0,

(4.27)

i=1

then one has the inequality "

n X i=1

2

wi |xi |

n X

# 21 2

wi |yi |

i=1

n

1 M +mX ≤ · √ wi Re (xi y¯i ) 2 mM i=1 n 1 M + m X ≤ · √ wi xi y¯i . 2 mM

(4.28)

i=1

The following corollary also holds. Corollary 150 With the assumptions in Corollary 149, then one has the following inequality: " n # 21 n n X X X 0≤ wi |xi |2 wi |yi |2 − wi xi y¯i (4.29) i=1 i=1 i=1 1 " n # n n 2 X X X ≤ wi |xi |2 wi |yi |2 − wi Re (xi y¯i ) i=1

i=1

√

√ 2 M− m √ ≤ 2 mM √ √ 2 M− m √ ≤ 2 mM

i=1 n X

wi Re (xi y¯i )

i=1

n X wi xi y¯i i=1

and n 2 n X X 0≤ wi |xi |2 wi |yi |2 − wi xi y¯i i=1 i=1 i=1 " n #2 n n X X X 2 2 ≤ wi |xi | wi |yi | − wi Re (xi y¯i ) n X

i=1

i=1

i=1

(4.30)

110

CHAPTER 4. COUNTERPART INEQUALITIES " n #2 (M − m)2 X ≤ wi Re (xi y¯i ) 4mM i=1 2 n 2 X (M − m) ≤ wi xi y¯i . 4mM i=1

4.5

A Counterpart Inequality for Real Numbers

The following result holds [10, Proposition 5.1]. Theorem 151 Let a, A ∈ R and x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) be two sequences with the property that: ayi ≤ xi ≤ Ayi for each i ∈ {1, . . . , n} .

(4.31)

Then for any w ¯ = (w1 , . . . , wn ) a sequence of positive real numbers, one has the inequality !2 n n n X X X 0≤ wi x2i wi yi2 − wi xi yi (4.32) i=1

≤ The constant

1 4

i=1

i=1 n X

1 (A − a)2 4

!2 wi yi2

.

i=1

is sharp in (4.32).

Proof. Let us define I1 :=

A

n X

wi yi2

−

i=1

and I2 :=

n X

! wi xi yi

n X

i=1 n X

! wi yi2

wi xi yi − a

i=1 n X

n X

! wi yi2

i=1

(Ayi − xi ) (xi − ayi ) wi .

i=1

i=1

Then I1 = (a + A)

n X i=1

wi yi2

n X i=1

wi xi yi −

n X i=1

!2 wi xi yi

− aA

n X i=1

!2 wi yi2

4.5. A COUNTERPART INEQUALITY FOR REAL NUMBERS

111

and I2 = (a + A)

n X

wi yi2

i=1

n X

wi xi yi −

n X

i=1

wi x2i

i=1

n X

wi yi2

− aA

n X

i=1

!2 wi yi2

i=1

giving I1 − I2 =

n X

wi x2i

i=1

n X

n X

wi yi2 −

i=1

!2 wi yi2

.

(4.33)

i=1

If (4.31) holds, then (Ayi − xi ) (xi − ayi ) ≥ 0 for each i ∈ {1, . . . , n} and thus I2 ≥ 0 giving !2 n n n X X X wi x2i wi yi2 − wi yi2 i=1

i=1 "

≤

i=1

A

n X

wi yi2 −

i=1

n X

! wi xi yi

i=1

n X

wi xi yi − a

i=1

n X

!# wi yi2

. (4.34)

i=1

If we use the elementary inequality for real numbers u, v ∈ R uv ≤

1 (u + v)2 , 4

(4.35)

then we have for u := A

n X

wi yi2 −

i=1

n X

wi xi yi ,

v :=

n X

i=1

wi xi yi − a

i=1

n X

wi yi2

i=1

that A

n X

wi yi2 −

i=1

n X

! wi xi yi

i=1

n X i=1

wi xi yi − a

n X

! wi yi2

i=1

1 ≤ (A − a)2 4

n X i=1

and the inequality (4.32) is proved. Now, assume that (4.32) holds with a constant c > 0, i.e., !2 !2 n n n n X X X X 2 wi x2i wi yi2 − wi xi yi ≤ c (A − a) wi yi2 , i=1

i=1

i=1

i=1

!2 wi yi2

(4.36)

112

CHAPTER 4. COUNTERPART INEQUALITIES

where a, A, x ¯, y ¯ satisfy (4.31). We choose n = 2, w1 = w2 = 1 and let a, A, y1 , y2 , x, α ∈ R such that ay1 < x1 = Ay1 , ay2 = x2 < Ay2 . With these choices, we get from (4.36) that
2
2

a2 y12 + a2 y22 y12 + y22 − A2 y12 + a2 y22 ≤ c (A − a)2 y12 + y22 , which is equivalent to (A − a)2 y12 y22 ≤ c (A − a)2 y12 + y22


2

.

Since we may choose a 6= A, we deduce y12 y22 ≤ c y12 + y22


2

,

giving, for y1 = y2 = 1, c ≥ 41 . The following corollary is obvious. Corollary 152 With the above assumptions for a, A, x ¯ and y ¯, we have the inequality !2 !2 n n n n X X X X 1 2 0≤ x2i yi2 − xi yi ≤ (A − a) yi2 . (4.37) 4 i=1 i=1 i=1 i=1 Remark 153 Condition (4.31) may be replaced by the weaker condition n X

wi (Ayi − xi ) (xi − ayi ) ≥ 0

(4.38)

i=1

and the conclusion in Theorem 151 will still be valid, i.e., the inequality (4.32) holds. For (4.37) to be true it suffices that n X i=1

holds true.

(Ayi − xi ) (xi − ayi ) ≥ 0

(4.39)

4.6. A COUNTERPART INEQUALITY FOR COMPLEX NUMBERS 113

4.6

A Counterpart Inequality for Complex Numbers

The following result holds [10, Proposition 5.1]. Theorem 154 Let a, A ∈ C and x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) ∈ Cn , w ¯ = (w1 , . . . , wn ) ∈ Rn+ . If n X

wi Re [(Ayi − xi ) (¯ xi − a ¯y¯i )] ≥ 0,

(4.40)

i=1

then one has the inequality 2 n n X X 2 2 0≤ wi |xi | wi |yi | − wi xi y¯i i=1 i=1 i=1 ! 2 n X 1 2 2 wi |yi | . ≤ |A − a| 4 i=1 n X

The constant

1 4

(4.41)

is sharp in (4.41).

Proof. Consider " ! n !# n n n X X X X A1 := Re A wi |yi |2 − wi xi y¯i wi x¯i yi − a ¯ wi |yi |2 i=1

i=1

and A2 :=

n X

i=1

" wi |yi |2 − Re

i=1

n X

i=1

# wi (Ayi − xi ) (¯ xi − a ¯y¯i ) .

i=1

Then A1 =

n X

" wi |yi |2 − Re A

i=1

n X

wi x¯i yi + a ¯

i=1

n X

# wi xi y¯i

i=1

2 !2 n n X X 2 − wi xi y¯i − Re (¯ aA) wi |yi | i=1

i=1

114

CHAPTER 4. COUNTERPART INEQUALITIES

and A2 =

n X

" wi |yi |2 − Re A

n X

i=1

−

n X

wi x¯i yi + a ¯

n X

i=1

wi |xi |2

i=1

n X

# wi xi y¯i

i=1 n X

wi |yi |2 − Re (¯ aA)

i=1

!2 wi |yi |2

i=1

giving A 1 − A2 =

n X i=1

n 2 n X X 2 2 wi |xi | wi |yi | − wi xi y¯i . i=1

(4.42)

i=1

If (4.40) holds, then A2 ≥ 0 and thus n X i=1

n 2 n X X 2 2 wi |xi | wi |yi | − wi xi y¯i i=1 " i=1 n ! n X X ≤ Re A wi |yi |2 − wi xi y¯i i=1

i=1

×

n X

wi x¯i yi − a ¯

i=1

n X

!# wi |yi |2

. (4.43)

i=1

If we use the elementary inequality for complex numbers z, t ∈ C 1 Re [z t¯] ≤ |z − t|2 , (4.44) 4 Pn Pn Pn 2 then ¯i − i=1 wi xi y Pn we have2 for z := A i=1 wi |yi | − i=1 wi xi y¯i , t := a i=1 wi |yi | that " Re

A

n X i=1

wi |yi |2 −

n X

! wi xi y¯i

i=1

n X i=1

≤ and the inequality (4.41) is proved.

wi x¯i yi − a ¯

n X

!# wi |yi |2

i=1

1 |A − a|2 4

n X i=1

!2 wi |yi |2

(4.45)

4.6. A COUNTERPART INEQUALITY FOR COMPLEX NUMBERS 115 Now, assume that (4.41) holds with a constant c > 0, i.e., 2 !2 n n n n X X X X wi |xi |2 wi |yi |2 − wi xi y¯i ≤ c |A − a|2 wi |yi |2 , (4.46) i=1

i=1

i=1

i=1

where x ¯, y ¯, a, A satisfyP (4.40). P n Consider y ¯ ∈ C , ni=1 |yi |2 wi = 1, a 6= A, m ¯ ∈ Cn , ni=1 wi |mi |2 = 1 Pn with i=1 wi yi mi = 0. Define xi :=

A+a A+a yi + mi , 2 2

i ∈ {1, . . . , n} .

Then n X i=1

n A − a 2 X wi (Ayi − xi ) (¯ xi − a ¯ yi ) = wi (yi − mi ) (¯ yi − m ¯ i) = 0 2 i=1

and thus the condition (4.40) is fulfilled. From (4.46) we deduce 2 2  n n  X X A + a A − a A + a A − a y + m w − y + m y ¯ w i i i i i i i 2 2 2 2 i=1 i=1 ≤ c |A − a|2 and since n X i=1

and

  2 A+a A + a 2 A − a 2 A − a − wi yi + mi = 2 2 2 2 2  n  X A + a 2 A + a A − a yi + mi y¯i wi = 2 2 2 i=1

then by (4.46) we get |A − a|2 ≤ c |A − a|2 4 giving c ≥ 14 and the theorem is completely proved. The following corollary holds.

116

CHAPTER 4. COUNTERPART INEQUALITIES

Corollary 155 Let a, A ∈ C and x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) ∈ Cn be with the property that n X

Re [(Ayi − xi ) (¯ xi − a ¯y¯i )] ≥ 0,

(4.47)

i=1

then one has the inequality 0≤

n X i=1

2 n n X X 1 2 2 |xi | |yi | − xi y¯i ≤ |A − a|2 4 i=1 i=1

The constant

1 4

n X

!2 |yi |2

.

(4.48)

i=1

is best in (4.48).

Remark 156 A sufficient condition for both (4.40) and (4.47) to hold is Re [(Ayi − xi ) (¯ xi − a ¯y¯i )] ≥ 0

(4.49)

for any i ∈ {1, . . . , n} .

4.7

Shisha-Mond Type Inequalities

As some particular case for bounds on differences of means, O. Shisha and B. Mond obtained in 1967 (see [23]) the following counterpart of (CBS) − inequality: ¯ = (b1 , . . . , bn ) are such Theorem 157 Assume that ¯ a = (a1 , . . . , an ) and b that there exists a, A, b, B > 0 with the property that: a ≤ aj ≤ A and b ≤ bj ≤ B for any j ∈ {1, . . . , n}

(4.50)

then we have the inequality n X j=1

a2j

n X j=1

b2j −

n X j=1

!2 aj b j

r ≤

A − b

r !2 n n X X a aj b j b2j . B j=1 j=1

(4.51)

The equality holds in (4.51) if and only if there exists a subsequence (k1 , . . . , kp ) of (1, 2, . . . , n) such that   12   32 n A B =1+ , p a b akµ = A, bkµ = b (µ = 1, . . . , p) and ak = a, bk = B for every k distinct from all kµ .

4.7. SHISHA-MOND TYPE INEQUALITIES

117

Using another result stated for weighted means in [23], we may prove the following counterpart of the (CBS) −inequality. ¯ are positive sequences and there exists Theorem 158 Assume that ¯ a, b γ, Γ > 0 with the property that 0<γ≤

ai ≤ Γ < ∞ for any i ∈ {1, . . . , n} . bi

(4.52)

Then we have the inequality 0≤

n X i=1

a2i

n X

! 12 b2i

−

n X

i=1

i=1

n

(Γ − γ)2 X 2 ai b i ≤ b. 4 (γ + Γ) j=1 j

(4.53)

The equality holds in (4.53) if and only if there exists a subsequence (k1 , . . . , kp ) of (1, 2, . . . , n) such that p X m=1

n

b2km

Γ + 3γ X 2 akm ak = bj , = Γ (m = 1, . . . , p) and =γ 4 (γ + Γ) j=1 b km bk

for every k distinct from all km . Proof. In [23, p. 301], Shisha and Mond have proved the following weighted inequality 0≤

n X

! 21 qj x2j

−

n X

qj x j ≤

j=1

j=1

(C − c)2 , 4 (c + C)

(4.54)

P provided qj ≥ 0 (j = 1, . . . , n) with nj=1 qj = 1 and 0 < c ≤ xj < C < ∞ for any j ∈ {1, . . . , n} . Equality holds in (4.54) if and only if there exists a subsequence (k1 , . . . , kp ) of (1, 2, . . . , n) such that p X m=1

q km =

C + 3c , 4 (c + C)

xkm = C (m = 1, 2, . . . , p) and xk = c for every k distinct from all km .

(4.55)

118

CHAPTER 4. COUNTERPART INEQUALITIES

If in (4.54) we choose xj =

b2j aj , qj = Pn 2 , j ∈ {1, . . . , n} ; bj k=1 bk

then we get ! 12 Pn Pn 2 (Γ − γ)2 j=1 aj j=1 aj bj Pn 2 − Pn 2 ≤ , 4 (γ + Γ) k=1 bk k=1 bk giving the desired inequality (4.53). The case of equality follows by the similar case in (4.54) and we omit the details.

4.8

Zagier Type Inequalities

The following result was obtained by D. Zagier in 1995, [24]. Lemma 159 Let f, g : [0, ∞) → R be monotone decreasing nonnegative functions on [0, ∞). Then R∞ R∞ Z ∞ f (x) F (x) dx 0 g (x) G (x) dx 0 R ∞ , R∞ f (x) g (x) dx ≥ (4.56) max 0 F (x) dx, 0 G (x) dx 0 for any integrable functions F, G : [0, ∞) → [0, 1] . Proof. We will follow the proof in [24]. For all x ≥ 0 we have Z ∞ Z ∞ Z ∞ f (t) F (t) dt = f (x) F (t) dt + [f (t) − f (x)] F (t) dt 0 0 0 Z ∞ Z x ≤ f (x) F (t) dt + [f (t) − f (x)] dt 0

0

R∞ and hence, since 0 G (t) dt is bounded from above by both x and 0 G (t) dt, Z ∞ Z x f (t) F (t) dt · G (t) dt 0 0 Z ∞ Z ∞ Z x ≤ xf (x) F (t) dt + G (t) dt · [f (t) − f (x)] dt 0 0 0 Z ∞  Z x Z ∞ ≤ max F (t) dt, G (t) dt · f (t) dt. Rx

0

0

0

4.8. ZAGIER TYPE INEQUALITIES

119

Now, multiply R ∞ by −dg (x) and R ∞integrate by parts from 0 to ∞. The left hand side gives −∞ f (t) F (t) dt · −∞ g (t) G (t) dt, the right hand side gives Z ∞  Z ∞ Z ∞ max F (t) dt, G (t) dt · f (t) g (t) dt, 0

0

0

and the inequality remains true because the measure −dg (x) is nonnegative. The following particular case is a counterpart of the (CBS) −integral inequality obtained by D. Zagier in 1977, [25]. Corollary 160 If f, g : [0, ∞) → [0, ∞) are decreasing function on [0, ∞), then   Z ∞ Z ∞ Z ∞ max f (0) g (t) dt, g (0) f (t) dt · f (t) g (t) dt 0 0 0 Z ∞ Z ∞ 2 ≥ f (t) dt g 2 (t) dt. (4.57) 0

0

Remark 161 The following weighted version of (4.56) may be proved in a similar way, as noted by D. Zagier in [25] R∞ R∞ Z ∞ w (t) f (t) F (t) dt 0 w (t) f (t) G (t) dt 0 R ∞ . (4.58) R∞ w (t) f (t) g (t) dt ≥ max 0 w (t) F (t) dt, 0 w (t) G (t) dt 0 provided w (t) > 0 on [0, ∞), f, g : [0, ∞) → [0, ∞) are monotonic decreasing and F, G : [0, ∞) → [0, 1] are integrable on [0, ∞). We may state and prove the following discrete inequality. ¯= Theorem 162 Consider the sequences of real numbers ¯ a = (a1 , . . . , an ), b (b1 , . . . , bn ) , p ¯ = (p1 , . . . , pn ), q ¯ = (q1 , . . . , qn ) and w ¯ = (w1 , . . . , wn ) . If ¯ are decreasing and nonnegative; (i) ¯ a and b (ii) pi , qi ∈ [0, 1] and wi ≥ 0 for any i ∈ {1, . . . , n} , then we have the inequality n X i=1

Pn

wi ai bi ≥

P wi pi ai ni=1 wi qi bi i=1 P P . max { ni=1 wi pi , ni=1 wi qi }

(4.59)

120

CHAPTER 4. COUNTERPART INEQUALITIES

Proof. Consider the functions f, g, F, G, W : [0, ∞) → R given by     a , t ∈ [0, 1) b1 , t ∈ [0, 1) 1         a , t ∈ [1, 2)  2  b2 , t ∈ [1, 2) . .. .. f (t) = , g (t) = , .       an , t ∈ [n − 1, n) bn , t ∈ [n − 1, n)      0  0 t ∈ [n, ∞) t ∈ [n, ∞)     p1 , t ∈ [0, 1) q1 , t ∈ [0, 1)         p , t ∈ [1, 2)  2  q2 , t ∈ [1, 2) . .. .. F (t) = , G (t) = , .       pn , t ∈ [n − 1, n) qn , t ∈ [n − 1, n)      0  0 t ∈ [n, ∞) t ∈ [n, ∞) and

  w1 ,      w2 , .. W (t) = .    wn ,    0

t ∈ [0, 1) t ∈ [1, 2) . t ∈ [n − 1, n) t ∈ [n, ∞)

We observe that, the above functions satisfy the hypothesis of Remark 161 and since, for example, Z ∞ Z ∞ n Z i X w (t) f (t) g (t) dt = w (t) f (t) g (t) dt + w (t) f (t) g (t) dt 0

=

i=1 n X

i−1

n

wk ak bk ,

k=1

then by (4.58) we deduce the desired inequality (4.59). Remark 163 A similar inequality for sequences under some monotonicity assumptions for p ¯ and q ¯ was obtained in 1995 by J. Peˇcari´c in [26]. The following counterpart of the (CBS) −discrete inequality holds. ¯ are decreasing nonnegative sequences with Theorem 164 Assume that ¯ a, b a1 , b1 6= 0 and w ¯ a nonnegative sequence. Then ( n ) n n n n X X X X X wi a2i wi b2i ≤ max b1 wi ai , a1 wi bi wi ai bi . (4.60) i=1

i=1

i=1

i=1

i=1

4.9. A COUNTERPART IN TERMS OF THE SUP −NORM The proof follows by Theorem 162 on choosing pi = [0, 1] , i ∈ {1, . . . , n} . We omit the details.

ai a1

121

∈ [0, 1] , qi =

bi b1

∈

Remark 165 When wi = 1, we recapture Alzer’s result from 1992, [27].

4.9

A Counterpart in Terms of the sup −Norm

The following result has been proved in [11]. Lemma 166 Let α ¯ = (α1 , . . . , αn ) and x ¯ = (x1 , . . . , xn ) be sequences of complex numbers and p ¯ = (p1 , . . . , pn ) a sequence of nonnegative real numbers P such that ni=1 pi = 1. Then one has the inequality n n n X X X pi α i xi − pi α i pi xi i=1

i=1

i=1

 n X  i2 p i − ≤ max |∆αi | max |∆xi | i=1,n−1

i=1,n−1

i=1

n X

!2  ipi

 , (4.61)

i=1

where ∆αi is the forward difference, i.e., ∆αi := αi+1 − αi . Inequality (4.61) is sharp in the sense that the constant C = 1 in the right membership cannot be replaced be a smaller one. Proof. We shall follow the proof in [11]. We start with the following identity n X i=1

pi α i xi −

n X

pi α i

i=1

n X i=1

n 1X pi pj (αi − αj ) (xi − xj ) pi xi = 2 i,j=1 X = pi pj (αi − αj ) (xi − xj ) . 1≤i<j≤n

As i < j, we can write that αj − αi =

j−1 X k=i

∆αk

122

CHAPTER 4. COUNTERPART INEQUALITIES

and xj − xi =

j−1 X

∆xk .

k=i

Using the generalised triangle inequality, we have successively n n n X X X pi α i xi − pi α i pi xi i=1 i=1 i=1 j−1 j−1 X X X = pi pj ∆αk ∆xk 1≤i<j≤n k=i k=i j−1 j−1 X X X ≤ pi pj ∆αk ∆xk 1≤i<j≤n

k=i

k=i

j−1

j−1

X

≤

1≤i<j≤n

pi pj

X

|∆αk |

k=i

X

|∆xk |

k=i

:= A. Note that |∆αk | ≤ max |∆αs | 1≤s≤n−1

and |∆xk | ≤ max |∆xs | 1≤s≤n−1

for all k = i, . . . , j − 1 and then by summation j−1 X k=i

|∆αk | ≤ (j − i) max |∆αs | 1≤s≤n−1

and j−1 X k=i

|∆xk | ≤ (j − i) max |∆xs | . 1≤s≤n−1

Taking into account the above estimations, we can write " # X A≤ pi pj (j − i)2 max |∆αs | max |∆xs | . 1≤i<j≤n

1≤s≤n−1

1≤s≤n−1

4.9. A COUNTERPART IN TERMS OF THE SUP −NORM

123

As a simple calculation shows that X

pi pj (j − i)2 =

n X

i2 p i −

i=1

1≤i<j≤n

n X

!2 ipi

,

i=1

inequality (4.61) is proved. To prove the sharpness of the constant, let us assume that (4.61) holds with a constant C > 0, i.e., n n n X X X pi α i xi − pi α i pi xi i=1

i=1

i=1

 n X  ≤ C max |∆αi | max |∆xi | i2 p i − i=1,n−1

i=1,n−1

i=1

n X

!2  ipi

 . (4.62)

i=1

Now, choose the sequences αk = α + kβ (β 6= 0) and xk = x + ky (y 6= 0) , k ∈ {1, . . . , n} to get n n n n X 1 X X X pi α i xi − pi α i pi xi = pi pj (i − j) βy 2 i,j=1 i=1 i=1 i=1  n X = |β| |y|  i2 p i − i=1

n X

!2  ipi



i=1

and  n X  max |∆αi | max |∆xi | i2 p i −

i=1,n−1

i=1,n−1

i=1

n X

!2  ipi



i=1

 n X  = |β| |y| i2 p i − i=1

n X i=1

and then, by (4.62), we get C ≥ 1. The following counterpart of the (CBS) −inequality holds [12].

!2  ipi



124

CHAPTER 4. COUNTERPART INEQUALITIES

¯ = (b1 , . . . , bn ) be two sequences of Theorem 167 Let ¯ a = (a1 , . . . , an ) and b real numbers with ai 6= 0, (i = 1, . . . , n) . Then one has the inequality 0≤

n X

a2i

i=1

n X

b2i −

i=1

n X

!2 ai b i

i=1

   2 X n n X bk 2  ≤ max ∆ ai i2 a2i − ak k=1,n−1 i=1 i=1

n X

!2  ia2i  .

i=1

The constant C = 1 is sharp in the sense that it cannot be replaced by a smaller constant. Proof. Follows by Lemma 166 on choosing a2 pi = Pn i

2 k=1 ak

, αi =

bi bi , xi = , i ∈ {1, . . . , n} ai ai

and performing some elementary calculations. We omit the details.

4.10

A Counterpart in Terms of the 1−Norm

The following result has been obtained in [13]. Lemma 168 Let α ¯ = (α1 , . . . , αn ) and x ¯ = (x1 , . . . , xn ) be sequences of complex numbers and p ¯ = (p1 , . . . , pn ) a sequence of nonnegative real numbers P such that ni=1 pi = 1. Then one has the inequality n n n X X X p α x − p α p x i i i i i i i i=1

i=1

i=1

n

n−1

n−1

X X 1X ≤ pi (1 − pi ) |∆αi | |∆xi | , (4.63) 2 i=1 i=1 i=1 where ∆αi := αi+1 − αi is the forward difference. The constant 12 is sharp in the sense that it cannot be replaced by a smaller constant.

4.10. A COUNTERPART IN TERMS OF THE 1−NORM

125

Proof. We shall follow the proof in [13]. As in the proof of Lemma 166 in Section 4.9, we have n n n X X X pi α i xi − pi α i pi xi i=1

i=1

i=1

X

≤

pi pj

j−1 X

1≤i<j≤n

|∆αk |

k=i

j−1 X

|∆xl | := A. (4.64)

l=i

It is obvious that for all 1 ≤ i < j ≤ n − 1, we have that j−1 X

and

|∆αk | ≤

n−1 X

k=i

k=1

j−1 X

n−1 X

|∆xl | ≤

l=i

|∆αk |

|∆xl | .

l=1

Utilising these and the definition of A, we conclude that A≤

n−1 X

|∆αk |

k=1

n−1 X

|∆xl |

l=1

X

pi pj .

(4.65)

1≤i<j≤n

Now, let us observe that " n # X 1 X pi pj − pi pj pi pj = 2 i,j=1 i=j 1≤i<j≤n " n # n n X 1 X X = pi pj − p2i 2 i=1 j=1 i=1 X

(4.66)

n

=

1X pi (1 − pi ) . 2 i=1

Making use of (4.64) – (4.66), we deduce the desired inequality (4.63). To prove the sharpness of the constant 12 , let us assume that (4.63) holds with a constant C > 0. That is n n n n n−1 n−1 X X X X X X pi α i xi − pi α i pi xi ≤ C pi (1 − pi ) |∆αi | |∆xi | (4.67) i=1

i=1

i=1

i=1

i=1

i=1

126

CHAPTER 4. COUNTERPART INEQUALITIES

for all αi , xi , pi (i = 1, . . . , n) as above and n ≥ 1. Choose in (4.63) n = 2 and compute 2 X

pi α i xi −

i=1

2 X

2 X

pi α i

i=1

i=1

2 1X pi xi = pi pj (αi − αj ) (xi − xj ) 2 i,j=1 X = pi pj (αi − αj ) (xi − xj ) 1≤i<j≤2

= p1 p2 (α1 − α2 ) (x1 − x2 ) . Also 2 X i=1

pi (1 − pi )

2 X

|∆αi |

i=1

2 X

|∆xi | = (p1 p2 + p1 p2 ) |α1 − α2 | |x1 − x2 | .

i=1

Substituting in (4.67), we obtain p1 p2 |α1 − α2 | |x1 − x2 | ≤ 2Cp1 p2 |α1 − α2 | |x1 − x2 | . If we assume that p1 , p2 > 0, α1 6= α2 , x1 6= x2 , then we obtain C ≥ 12 , which proves the sharpness of the constant 12 . We are now able to state the following counterpart of the (CBS) −inequality [12]. ¯ = (b1 , . . . , bn ) be two sequences of Theorem 169 Let ¯ a = (a1 , . . . , an ) and b real numbers with ai 6= 0 (i = 1, . . . , n) . Then one has the inequality !2 n n n X X X 0≤ a2i b2i − ai b i (4.68) i=1

i=1

i=1

" n−1   #2 X X ∆ bk ≤ a2i a2j . ak 1≤i<j≤n

k=1

The constant C = 1 is sharp in (4.68), in the sense that it cannot be replaced by a smaller constant. Proof. We choose a2i P pi = n

2 k=1 ak

,

α i = xi =

bi , i ∈ {1, . . . , n} ai

4.11. A COUNTERPART IN TERMS OF THE P −NORM

127

in (4.63) to get Pn 2 P 2 bi ( ni=1 ai bi ) i=1 0 ≤ Pn − P 2 2 ( nk=1 a2k ) k=1 ak  Pn 2  a2 Pn i 2 a 1 − i=1 i 1 k=1 ak Pn ≤ · 2 2 k=1 ak 1 = · 2

 !2 n−1  X b j ∆ aj j=1

 n−1  X ∆ bj aj

Pn

Pn 2 a2k − a2i ) i=1 ai ( Pn k=12 2 ( k=1 ak )

!2

j=1

which is clearly equivalent to 0≤

n X

a2i

i=1

 1 ≤  2 Since

 1 2

n X

b2i −

i=1 n X

k=1

ai b i

i=1

!2 a2k

−

n X i=1

k=1

n X

!2

n X

!2 a2k

−



 !2 n  X b ∆ j . a4i  aj j=1

n X

 X

a4i  =

i=1

a2i a2j

1≤i<j≤n

the inequality (4.68) is thus proved.

4.11

A Counterpart in Terms of the p−Norm

The following result has been obtained in [14]. Lemma 170 Let α ¯ = (α1 , . . . , αn ) and x ¯ = (x1 , . . . , xn ) be sequences of complex numbers and p ¯ = (p , . . . , p ) a sequence of nonnegative real numbers 1 n Pn such that i=1 pi = 1. Then one has the inequality n n n X X X pi α i xi − pi α i pi xi i=1 i=1 i=1 ! p1 n−1 ! 1q n−1 X X X q p ≤ (i − j) pi pj |∆xk | , (4.69) |∆αk | 1≤j
k=1

k=1

128

CHAPTER 4. COUNTERPART INEQUALITIES

where p > 1, p1 + 1q = 1. The constant C = 1 in the right hand side of (4.69) is sharp in the sense that it cannot be replaced by a smaller constant. Proof. We shall follow the proof in [14]. As in the proof of Lemma 166 in Section 4.9, we have n n n X X X pi α i xi − pi α i pi xi i=1

i=1

i=1

≤

X

pi pj

1≤j
i−1 X

|∆αk |

k=j

i−1 X

|∆xl | := A. (4.70)

l=j

Using H¨older’s discrete inequality, we can state that i−1 X

i−1 X

1

|∆αk | ≤ (i − j) q

k=j

k=j

i−1 X

i−1 X

! p1 |∆αk |p

and 1

|∆xl | ≤ (i − j) p

l=j

where p > 1,

A≤

1 p

+

1 q

X

,

= 1, and then we get

pi pj (i − j)

i−1 X

! p1 |∆αk |p

k=j

i−1 X

|∆αk |p ≤

k=j

and

|∆xl |q

l=j

1≤j
Since

! 1q

i−1 X k=j

i−1 X

! 1q |∆xk |q

.

(4.71)

k=j

n−1 X

|∆αk |p

k=1

q

|∆xk | ≤

n−1 X

|∆xk |q .

k=1

for all 1 ≤ j < i ≤ n, then by (4.70) and (4.71) we deduce the desired inequality (4.69).

4.11. A COUNTERPART IN TERMS OF THE P −NORM

129

To prove the sharpness of the constant, let us assume that (4.69) holds with a constant C > 0. That is, n n n X X X pi α i xi − pi α i pi xi i=1 i=1 i=1 ! p1 n−1 ! 1q n−1 X X X ≤C (i − j) pi pj |∆αk |p |∆xk |q . (4.72) 1≤j
k=1

k=1

Note that, for n = 2, we have 2 2 2 X X X pi α i xi − pi α i pi xi = p1 p2 |α1 − α2 | |x1 − x2 | i=1

i=1

i=1

and X

(i − j) pi pj

1≤j
1 X

! p1 |∆αk |p

k=1

1 X

! 1q |∆xk |q

k=1

= p1 p2 |α1 − α2 | |x1 − x2 | . Therefore, from (4.72), we obtain p1 p2 |α1 − α2 | |x1 − x2 | ≤ Cp1 p2 |α1 − α2 | |x1 − x2 | for all α1 6= α2 , x1 6= x2 , p1 p2 > 0, giving C ≥ 1. We are able now to state the following counterpart of the (CBS) −inequality. ¯ = (b1 , . . . , bn ) be two sequences of Theorem 171 Let ¯ a = (a1 , . . . , an ) , b real numbers with ai 6= 0, (i = 1, . . . , n) . Then one has the inequality !2 n n n X X X ai b i 0≤ a2i b2i − i=1

≤

i=1

i=1

 p ! p1 n−1  X b ∆ k ak k=1

 q ! 1q n−1  X b ∆ k ak k=1

X 1≤j
where p > 1, p1 + 1q = 1. The constant C = 1 is sharp in the above sense.

(i − j) a2i a2j ,

130

CHAPTER 4. COUNTERPART INEQUALITIES

Proof. Follows by Lemma 170 for a2 pi = Pn i

2 k=1 ak

, α i = xi =

bi , i ∈ {1, . . . , n} . ai

The following corollary is a natural consequence of Theorem 171 for p = q = 2. ¯ we have Corollary 172 With the assumptions of Theorem 171 for ¯ a and b, 0≤ ≤

n X

i=1 n−1 X k=1

4.12

a2i

n X

∆

b2i −

i=1



n X

!2 ai b i

i=1

 2 bk X (i − j) a2i a2j . ak 1≤j
A Counterpart Via an Andrica-Badea Result

The following result is due to Andrica and Badea [15, p. 16]. Lemma 173 Let x ¯ = (x1 , . . . , xn ) ∈ I n = [m, M ]n be a sequence of real numbers and let S be the subset of {1, . . . , n} that minimises the expression X 1 p i − Pn , (4.73) 2 i∈S

P where Pn := ni=1 pi > 0, p ¯ = (p1 , . . . , pn ) is a sequence of nonnegative real numbers. Then  !2  n n X X 1 1 maxn  pi x2i − pi xi  x ¯∈I Pn i=1 Pn i=1 ! X X (M − m)2 = p i Pn − pi . (4.74) 2 P n i∈S i∈S

4.12. A COUNTERPART VIA AN ANDRICA-BADEA RESULT

131

Proof. We shall follow the proof in [15, p. 161]. Define n n 1 X 2 1 X Dn (¯ x, p ¯ ) := pi xi − pi xi Pn i=1 Pn i=1 1 X = pi pj (xi − xj )2 . Pn 1≤i<j≤n

!2

Keeping in mind the convexity of the quadratic function, we have Dn (α¯ x + (1 − α) y ¯, p ¯) X 1 = 2 pi pj [αxi + (1 − α) yi − αxj − (1 − α) yj ]2 Pn 1≤i<j≤n 1 X pi pj [α (xi − xj ) + (1 − α) (yi − yj )]2 = 2 Pn 1≤i<j≤n   1 X ≤ 2 pi pj α (xi − xj )2 + (1 − α) (yi − yj )2 Pn 1≤i<j≤n = αDn (¯ x, p ¯ ) + (1 − α) Dn (¯ y, p ¯) , hence Dn (·, p ¯ ) is a convex function on I n . Using a well known theorem (see for instance [16, p. 124]), we get that the maximum
of Dn (·, p ¯ ) is attained on the boundary of I n . Let S, S¯ be the partition of {1, . . . , n} such that the maximum of Dn (·, p ¯ ) is obtained for x ¯0 = (x01 , . . . , x0n ) , where x0i = m if i ∈ S¯ and x0i = M if i ∈ S. In this case we have Dn (¯ x0 , p ¯) =

1 Pn2

X

pi pj (xi − xj )2

1≤i<j≤n

! X (M − m)2 X = p i Pn − pi . Pn2 i∈S i∈S The expression ! X i∈S

p i Pn −

X i∈S

pi

(4.75)

132

CHAPTER 4. COUNTERPART INEQUALITIES

is a maximum when the set S minimises the expression X 1 p i − Pn . 2 i∈S

From (4.75) it follows that Dn (¯ x, p ¯ ) is also a maximum and the proof of the above lemma is complete. The following counterpart of the (CBS) −inequality holds. ¯ = (b1 , . . . , bn ) be two sequences of Theorem 174 Let ¯ a = (a1 , . . . , an ) and b real numbers with ai 6= 0 (i = 1, . . . , n) and bi ≤ M < ∞ for each i ∈ {1, . . . , n} . (4.76) −∞ < m ≤ ai Let S be the subset of {1, . . . , n} that minimizes the expression n X X 1 a2i , (4.77) a2i − 2 i=1

i∈S

and denote S¯ := {1, . . . , n} \ S. Then we have the inequality !2 n n n X X X 0≤ a2i b2i − ai b i i=1

i=1 2

≤ (M − m)

(4.78)

i=1

X

a2i

X i∈S¯

i∈S

n X

1 ≤ (M − m)2 4

a2i !2

a2i

.

i=1

Proof. The proof of the second inequality in (4.78) follows by Lemma 173 on choosing pi = a2i , xi = abii , i ∈ {1, . . . , n} . The third inequality is obvious as ! n X X X X X a2i a2i = a2i a2j − a2i i∈S

i∈S¯

j=1

i∈S

1 ≤ 4 =

1 4

X

a2i +

j=1

i∈S n X j=1

i∈S n X

!2 a2j

.

!2 a2j −

X i∈S

a2i

4.13. A REFINEMENT OF CASSELS’ INEQUALITY

4.13

133

A Refinement of Cassels’ Inequality

In 1914, P. Schweitzer [18] proved the following result. Theorem 175 If ¯ a = (a1 , . . . , an ) is a sequence of real numbers such that 0 < m ≤ ai ≤ M < ∞ (i ∈ {1, . . . , n}) , then ! ! n n 1X 1X 1 (M + m)2 ai ≤ . (4.79) n i=1 n i=1 ai 4mM In 1972, A. Lupa¸s [17] proved the following refinement of Schweitzer’s result which gives the best bound for n odd as well. Theorem 176 With the assumptions in Theorem 175, one has n  n+1 
 n+1  n
n n X X M + m M + m 1 2 2 2 ai ≤ 2 , a Mm i=1 i=1 i

(4.80)

where [·] is the integer part. In 1988, Andrica and Badea [15] established a weighted version of Schweitzer and Lupa¸s inequalities via the use of the following weighted version of the Gr¨ uss inequality [15, Theorem 2]. Theorem 177 If m1 ≤ ai ≤ M1 , m2 ≤ bi ≤ M2 (i ∈ {1, . . . , n}) and S is the subset of {1, . . . , n} which minimises the expression X 1 (4.81) p i − Pn , 2 i∈S P where Pn := ni=1 pi > 0, then n n n X X X p i ai b i − p i ai · p i bi (4.82) Pn i=1 i=1 i=1 ! X X ≤ (M1 − m1 ) (M2 − m2 ) p i Pn − pi . i∈S

1 ≤ Pn2 (M1 − m1 ) (M2 − m2 ) . 4

i∈S

134

CHAPTER 4. COUNTERPART INEQUALITIES

Proof. Using the result in Lemma 173, Section 4.12, we have n 1 X 2 p i ai − Pn i=1

n 1 X p i ai Pn i=1

!2

X (M1 − m1 )2 X P − pi ≤ p i n Pn2 i∈S i∈S

! (4.83)

and n 1 X 2 p i bi − Pn i=1

n 1 X pi bi Pn i=1

!2

2

X (M2 − m2 ) X P − pi ≤ p i n Pn2 i∈S i∈S

! (4.84)

and since !2 n n n 1 X 1 X 1 X p i ai b i − p i ai · p i bi Pn i=1 Pn i=1 Pn i=1  !2  n n 1 X 2 1 X ≤ p i ai − p i ai  Pn i=1 Pn i=1  !2  n n X X 1 1 × pi b2i − pi bi  , (4.85) Pn i=1 Pn i=1 the first part of (4.82) holds true. The second part follows by the elementary inequality 1 (a + b)2 , a, b ∈ R 4 P P for the choices a := i∈S pi , b := Pn − i∈S pi . We are now able to state and prove the result of Andrica and Badea [15, Theorem 4], which is related to Schweitzer’s inequality. ab ≤

Theorem 178 If 0 < m ≤ ai ≤ M < ∞, i ∈ {1, . . . , n} and S is a subset of {1, . . . , n} that minimises the expression X Pn pi − , 2 i∈S

4.13. A REFINEMENT OF CASSELS’ INEQUALITY

135

then we have the inequality ! n ! ! n 2 X X X 1 X (M − m) p i ai pi ≤ Pn2 + p i Pn − pi a Mm i i=1 i=1 i∈S i∈S

(4.86)

(M + m)2 2 ≤ Pn . 4M m Proof. We shall follow the proof in [15]. We obtain from Theorem 176 with bi = a1i , m1 = m, M1 = m, m2 = M1 , M2 = m1 , the following !   n n X X X 1 1 1 X 2 pi ai pi ≤ (M − m) − p i Pn − pi , Pn − ai m M i=1

i=1

i∈S

i∈S

that leads, in a simple manner, to (4.86). We may now prove the following counterpart for the weighted (CBS) − inequality that improves the additive version of Cassels’ inequality. ¯ = (b1 , . . . , bn ) be two sequences of posTheorem 179 Let ¯ a = (a1 , . . . , an ) , b itive real numbers with the property that 0<m≤

bi ≤ M < ∞ for each i ∈ {1, . . . , n} ai

(4.87)

and p ¯ = (p1 , . . . , pn ) a sequence of nonnegative real numbers such that Pn := P n i=1 pi > 0. If S is a subset of {1, . . . , n} that minimises the expression n X 1X p i ai b i − p i ai b i (4.88) 2 i=1 i∈S then one has the inequality n X i=1

pi a2i

n X i=1

pi b2i

−

n X

!2 p i ai b i

(4.89)

i=1

n X X (M − m)2 X ≤ p i ai b i p i ai b i − p i ai b i Mm i=1 i∈S i∈S !2 n (M − m)2 X ≤ p i ai b i . 4M m i=1

!

136

CHAPTER 4. COUNTERPART INEQUALITIES

Proof. Applying Theorem 178 for ai = xi , pi = qi xi we may deduce the inequality n X i=1

qi x2i

n X

qi −

i=1

n X

!2 qi x i

i=1

! n X X (M − m)2 X ≤ qi x i qi x i − qi xi , (4.90) Mm i=1 i∈S i∈S P provided qi ≥ 0, ni=1 qi > 0, 0 < m ≤ xi ≤ M < ∞, for i ∈ {1, . . . , n} and S is a subset of {1, . . . , n} that minimises the expression Pn X q x i i i=1 qi x i − (4.91) . 2 i∈S

Now, if in (4.90) we choose qi = pi a2i , xi = abii ∈ [m, M ] for i ∈ {1, . . . , n} , we deduce the desired result (4.89). The following corollary provides a refinement of Cassels’ inequality. Corollary 180 With the assumptions of Theorem 179, we have the inequality Pn Pn 2 2 i=1 pi ai i=1 pi bi 1≤ (4.92) Pn 2 ( i=1 pi ai bi ) P P   (M − m)2 i∈S pi ai bi i∈S pi ai bi ≤1+ · Pn 1 − Pn Mm i=1 pi ai bi i=1 pi ai bi ≤

(M + m)2 . 4M m

The case of the “unweighted” Cassels’ inequality is embodied in the following corollary as well. ¯ satisfy (4.88). If S is a subset of Corollary 181 Assume that ¯ a and b {1, . . . , n} that minimises the expression n X X 1 ai b i − ai b i (4.93) 2 i=1 i∈S

4.14. TWO COUNTERPARTS VIA DIAZ-METCALF RESULTS then one has the inequality Pn 2 Pn 2 ai i=1 bi 1 ≤ i=1 Pn 2 ( i=1 ai bi )

137

(4.94)

P P   a b (M − m)2 i i i∈S i∈S ai bi ≤1+ · Pn 1 − Pn Mm i=1 ai bi i=1 ai bi ≤

(M + m)2 . 4M m

In particular, we may obtain the following refinement of the P´olya-Szeg¨o’s inequality. Corollary 182 Assume that 0 < a ≤ ai ≤ A < ∞, 0 < b ≤ bi ≤ B < ∞ for i ∈ {1, . . . , n} .

(4.95)

If S is a subset of {1, . . . , n} that minimises the expression (4.93), then one has the inequality Pn 2 Pn 2 ai i=1 bi (4.96) 1 ≤ i=1 Pn 2 ( i=1 ai bi ) P P   (AB − ab)2 i∈S ai bi i∈S ai bi ≤1+ · Pn 1 − Pn abAB i=1 ai bi i=1 ai bi (AB + ab)2 ≤ . 4abAB

4.14

Two Counterparts Via Diaz-Metcalf Results

In [19], J.B. Diaz and F.T. Metcalf proved the following inequality for sequences of complex numbers. ¯ = (b1 , . . . , bn ) be sequences of comLemma 183 Let ¯ a = (a1 , . . . , an ) and b plex numbers such that ak 6= 0, k ∈ {1, . . . , n} and         bk bk bk bk m ≤ Re + Im ≤ M, m ≤ Re − Im ≤ M, (4.97) ak ak ak ak

138

CHAPTER 4. COUNTERPART INEQUALITIES

where m, M ∈ R and k ∈ {1, . . . , n} . Then one has the inequality " n # n n X X X |bk |2 + mM |ak |2 ≤ (m + M ) Re ak¯bk k=1

k=1

(4.98)

k=1 n X ≤ |M + m| ak¯bk . k=1

Using the above result we may state and prove the following counterpart inequality. ¯ are as in (4.97) and m, M > 0, then one has the Theorem 184 If ¯ a and b inequality !2 n n n X X X (M + m)2 2 2 Re ak¯bk (4.99) |ak | |bk | ≤ 4mM k=1 k=1 k=1 2 n (M + m)2 X ¯ ≤ ak b k . 4mM k=1 Proof. Using the elementary inequality αp2 +

1 2 q ≥ 2pq, α > 0, p, q ≥ 0 α

we have √

n X

n n n X X 1 X 2 2 |bk | ≥ 2 |ak | |bk |2 mM |ak | + √ mM k=1 k=1 k=1 k=1 2

! 12 .

(4.100)

On the other hand, by (4.98), we have " n # n n X X √ 1 X (M + m) √ Re ak¯bk |bk |2 + mM |ak |2 ≤ √ mM mM k=1 k=1 k=1 n X M +m ¯ √ ak b k . ≤ mM k=1

(4.101)

Combining (4.100) and (4.101), we deduce the desired result (4.99). The following corollary is a natural consequence of the above lemma.

4.14. TWO COUNTERPARTS VIA DIAZ-METCALF RESULTS

139

¯ and m, M satisfy the hypothesis of Theorem 184, Corollary 185 If ¯ a and b then ! 12 n n n X X X 0≤ |ai |2 |bi |2 − ai¯bi (4.102) i=1 i=1 i=1 ! ! 12 n n n X X X − Re ak¯bk ≤ |ai |2 |bi |2 i=1 i=1 k=1 √  √ 2 ! n M − m X ¯ √ ≤ ak b k Re 2 mM k=1 √  √ 2 n M − m X ¯ √ ≤ ak b k 2 mM k=1

and 2 n X 0≤ |ai | |bi | − ai¯bi i=1 i=1 i=1 ! 2 n n n X X X ≤ |ai |2 |bi |2 − Re ai¯bi i=1 i=1 i=1 ! 2 n X (M − m)2 ¯ ≤ ai b i Re 4mM i=1 2 n (M − m)2 X ¯ ≤ ai b i . 4mM n X

2

n X

2

(4.103)

(4.104)

(4.105)

(4.106)

i=1

Another result obtained by Diaz and Metcalf in [19] is the following one. ¯ m and M be complex numbers such that Lemma 186 Let ¯ a, b,     bk bk + Im ≤ Re (M ) + Im (M ) ; Re (m) + Im (m) ≤ Re ak ak     bk bk Re (m) − Im (m) ≤ Re − Im ≤ Re (M ) − Im (M ) ; ak ak

(4.107)

140

CHAPTER 4. COUNTERPART INEQUALITIES

for each k ∈ {1, . . . , n} . Then n X

¯ |bk |2 + Re mM

n
X

k=1

" |ak |2 ≤ Re (M + m)

k=1

n X

# ak¯bk

(4.108)

n k=1 X ≤ |M + m| ak¯bk . k=1

The following counterpart result for the (CBS) −inequality may be stated as well.
¯ > 0, Theorem 187 With the assumptions in Lemma 186, and if Re mM then we have the inequality: "

n X k=1

2

|ak |

n X

# 12 2

|bk |

k=1

  P Re (M + m) nk=1 ak¯bk ≤ 
 1 ¯ 2 2 Re mM P |M + m| nk=1 ak¯bk ≤ . 
 1 ¯ 2 2 Re mM

(4.109)

The proof is similar to the one in Theorem 184 and we omit the details. Remark 188 Similar additive versions may be stated. They are left as an exercise to the interested reader.

4.15

ˇ Some Counterparts Via the Cebyˇ sev Functional

For x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) two sequences of real numbers and Pn p ¯ = (p1 , . . . , pn ) a sequence of nonnegative real numbers with i=1 pi = 1, ˇ sev functional define the Cebyˇ Tn (¯ p; x ¯, y ¯) :=

n X i=1

pi xi yi −

n X i=1

pi xi

n X i=1

pi yi .

(4.110)

ˇ ˇ 4.15. SOME COUNTERPARTS VIA THE CEBY SEV FUNCTIONAL 141 For x ¯ and p ¯ as above consider the norms: k¯ xk∞ := max |xi | i=1,n

n X

k¯ xkp¯,α :=

! α1 pi |xi |α

, α ∈ [1, ∞).

i=1

The following result holds [20]. Theorem 189 Let x ¯, y ¯, p ¯ be as above and ¯ c = (c, . . . , c) a constant sequence with c ∈ R. Then one has the inequalities 0 ≤ |Tn (¯ p; x ¯, y ¯)|  k¯ y − y¯µ,p kp¯,1 · inf k¯ x − c¯k∞ ;   c∈R      k¯ y − y¯µ,p kp¯,β · inf k¯ x − c¯kp¯,α , α > 1, α1 + β1 = 1; ≤ c∈R        k¯ y − y¯µ,p k∞ · inf k¯ x − c¯kp¯,1 ; c∈R   k¯ y − y¯µ,p kp¯,1 min k¯ xk∞ , k¯ x − x¯µ,p k∞ ;       n o  k¯ y − y¯µ,p kp¯,β min k¯ xkp¯,α , k¯ x − x¯µ,p kp¯,α , ≤   α > 1, α1 + β1 =o1;   n    k¯ y − y¯µ,p k · min k¯ xk , k¯ x − x¯µ,p k ; ∞

where xµ,p :=

p ¯ ,1

n X i=1

pi xi ,

yµ,p :=

(4.111)

p ¯ ,1

n X

pi yi

i=1

and x ¯µ , y ¯µ are the sequences with all components equal to xµ,p , yµ,p . Proof. Firstly, let us observe that for any c ∈ R, one has Sonin’s identity Tn (¯ p; x ¯, y ¯) = Tn (¯ p;¯ x − c¯, y¯ − y¯µ,p ) ! n n X X = pi (xi − c) yi − pj yj . i=1

j=1

Taking the modulus and using H¨older’s inequality, we have n X |Tn (¯ p; x ¯, y ¯)| ≤ pi |xi − c| |yi − yµ,p | i=1

(4.112)

(4.113)

142

CHAPTER 4. COUNTERPART INEQUALITIES P  max |xi − c| ni=1 pi (yi − yµ,p )   i=1,n       P 1 1 Pn α α n β β ≤ ( i=1 pi |xi − c| ) , i=1 pi |yi − yµ,p |    1 1  α > 1, α + β = 1;  Pn    i=1 pi |xi − c| max |yi − yµ,p | i=1,n  k¯ x − c¯k∞ k¯ y − y¯µ,p kp¯,1 ;      k¯ x − c¯kp¯,α k¯ y − y¯µ,p kp¯,β , α > 1, α1 + β1 = 1; =      k¯ x − c¯kp¯,1 k¯ y − y¯µ,p k∞ .

Taking the inf over c ∈ R in (4.113), we deduce the second inequality in (4.111). Since  xkp¯,α ,  k¯ inf k¯ x − c¯kp¯,α ≤ for any α ∈ [1, ∞] c∈R  k¯ x − x¯µ,p kp¯,α the last part of (4.110) is also proved. For p ¯ and x ¯ as above, define Tn (¯ p; x ¯) :=

n X

pi x2i −

i=1

n X

!2 pi xi

.

i=1

The following corollary holds [20]. Corollary 190 With the above assumptions we have 0 ≤ |Tn (¯ p; x ¯)|  x − x¯µ,p kp¯,1 · inf k¯ x − c¯k∞ ;   k¯ c∈R      k¯ x − x¯µ,p kp¯,β · inf k¯ x − c¯kp¯,α , α > 1, α1 + ≤ c∈R        k¯ x − x¯µ,p k∞ · inf k¯ x − c¯kp¯,1 ; c∈R

(4.114)

1 β

= 1;

ˇ ˇ 4.15. SOME COUNTERPARTS VIA THE CEBY SEV FUNCTIONAL 143   k¯ x − x¯µ,p kp¯,1 min k¯ xk∞ , k¯ x − x¯µ,p k∞ ;       o n  k¯ x − x ¯ k min k¯ x k , k¯ x − x ¯ k µ,p p µ,p p ≤ p ¯ ,α ¯ ,β ¯ ,α ,   α > 1, α1 + β1 =o1;   n    k¯ x − x¯µ,p k · min k¯ xk , k¯ x − x¯µ,p k . ∞

p ¯ ,1

p ¯ ,1

Remark 191 If pi := n1 , i = 1, . . . , n, then from Theorem 189 and Corollary 190 we recapture the results in [22]. The following counterpart of the (CBS) −inequality holds [20]. ¯ be two sequences of real numbers with ai 6= 0, i ∈ Theorem 192 Let ¯ a, b {1, . . . , n} . Then one has the inequality 0 ≤

n X i=1

a2i

n X

b2i

−

n X

i=1

!2 ai b i

(4.115)

i=1

 n " #  n X bi X a a ≤ inf max − c |ai | ak k i bk bi c∈R i=1,n ai i=1 k=1  bi   max  " # ai  n n X  i=1,n X ak ai × ≤ |ai | ak Pn bk bi   bi a b i=1 k=1  k k  .  max − Pk=1 n 2 a i=1,n ai k=1 k Proof. By Corollary 190, we may state that 0 ≤ Tn (¯ p; x ¯) ≤ k¯ x − x¯µ,p kp¯,1 · inf k¯ x − c¯k∞ c∈R  x k∞ ,  k¯ ≤ k¯ x − x¯µ,p kp¯,1 ×  k¯ x − x¯µ,p k∞ . For the choices a2i P pi = n

2 k=1 ak

, xi =

bi , i = 1, . . . , n; ai

(4.116)

144

CHAPTER 4. COUNTERPART INEQUALITIES

we get Pn Tn (¯ p; x ¯) =

k¯ x − x¯µ,p kp¯,1 = =

=

= =

k¯ x − c¯k∞

i=1

a2i

Pn

Pn 2 i=1 bi − ( i=1 Pn 2 2 ( k=1 ak )

2

ai b i )

,

n X pi xi − pj xj i=1 j=1 n n X X 1 1 2 bi Pn P ai − n aj b j 2 2 ai k=1 ak i=1 k=1 ak j=1 n n n X X X 1 2 2 a b a − a a b Pn i i j j k i 2 ( k=1 a2k ) i=1 j=1 k=1 n n X X 1 |a | a (a b − a b ) Pn i k k i i k 2 2 ( k=1 ak ) i=1 k=1 n n X X ak ai 1 , |ai | ak P 2 bk bi ( nk=1 a2k ) i=1 k=1

n X

bi bi xk∞ = max = max − c , k¯ i=1,n ai i=1,n ai

and k¯ x − x¯µ,p k∞

Pn b a b j j i j=1 = max − Pn . 2 a i=1,n ai k=1 k

Utilising the inequality (4.116) we deduce the desired result (4.115). The following result also holds [20]. Theorem 193 With the assumption in Theorem 192 and if α > 1, α1 + β1 = 1, then we have the inequality: 0≤

n X

a2i

i=1

 ≤

n X i=1

n X i=1

b2i −

n X

!2 ai b i

(4.117)

i=1

 β1 ! α1 β n n X X a a 2−β 2−α α |ai | |ai | |bi − cai | ak k i  inf bk bi c∈R 

k=1

i=1

ˇ ˇ 4.15. SOME COUNTERPARTS VIA THE CEBY SEV FUNCTIONAL 145  ≤

n X i=1

×

 β1 β n X a a |ai |2−β ak k i  bk bi 

k=1

   α1 n  P  2−α α  |ai | |bi |    i=1    1    Pn

k=1



Pn

i=1

a2k

2−α

|ai |

 1 ak ai α α Pn . k=1 ak bk bi

Proof. By Corollary 190, we may state that 0 ≤ Tn (¯ p; x ¯) ≤ k¯ x − x¯µ,p kp¯,β · inf k¯ x − c¯kp¯,α c∈R  xkp¯,α ,  k¯ ≤ k¯ x − x¯µ,p kp¯,β ×  k¯ x − x¯µ,p kp¯,α ,

(4.118)

for α > 1, α1 + β1 = 1. For the choices a2i P pi = n

k=1

, xi = 2

ak

bi , i = 1, . . . , n; ai

we get  k¯ x − x¯µ,p kp¯,β = 

n X i=1

 =

β  β1 n X pi xi − pj xj 

n X

j=1

a2  Pn i k=1 i=1

β  β1 Pn Pn 2 bi k=1 ak − ai j=1 aj bj  P a2k ai nk=1 a2k

1 = P 1+ 1 n ( k=1 a2k ) β

k¯ x − c¯kp¯,α =

n X i=1

! α1 pi |xi − c|α

 1 β β n n X X a a  |ai |2−β ak k i  , bk bi i=1

1 = P 1 n ( k=1 a2k ) α

k=1

n X i=1

! α1 |ai |2−α |bi − cai |α

,

146

CHAPTER 4. COUNTERPART INEQUALITIES n X

1 k¯ xkp¯,α = P 1 n ( k=1 a2k ) α

! α1 |ai |2−α |bi |α

i=1

and 1

k¯ x − x¯µ,p kp¯,α = P 1+ 1 ( nk=1 a2k ) α

n X i=1

α ! α1 n X a a |ai |2−α ak k i . bk bi k=1

Utilising the inequality (4.118), we deduce the desired result (4.117). Finally, the following result also holds [20]. Theorem 194 With the assumptions in Theorem 192 we have the following counterpart of the (CBS) −inequality: !2 n n n X X X 2 2 0≤ ai bi − ai bi (4.119) i=1

i=1 i=1 " n # n n b X X X i 2 ak − aj bj inf |ai | |bi − cai | ≤ max c∈R i=1,n ai j=1 i=1 k=1 n n b X X i ≤ max a2k − aj b j i=1,n ai j=1 k=1  n P   |ai bi |    i=1 ×  n n P ak ai  P 1  .  Pn |ai | ak  2 b b a k i k=1 k=1 k i=1

Proof. By Corollary 190, we may state that 0 ≤ Tn (¯ p; x ¯) ≤ k¯ x − x¯µ,p k∞ · inf k¯ x − c¯kp¯,1 c∈R  xkp¯,1 ,  k¯ ≤ k¯ x − x¯µ,p k∞  k¯ x − x¯µ,p kp¯,1 . For the choices a2i P pi = n

2 k=1 ak

, xi =

bi , i = 1, . . . , n; ai

(4.120)

¨ TYPE RESULT 4.16. ANOTHER COUNTERPART VIA A GRUSS

147

we get k¯ x − x¯µ,p k∞

Pn n b X i j=1 aj bj = max xi − pj xj = max − Pn 2 i=1,n ai i=1,n k=1 ak j=1 n n b X X 1 i 2 a − a b = Pn max j j , k 2 a a i k=1 k i=1,n j=1 k=1

k¯ x − c¯kp¯,1 =

n X

pi |xi − c| =

i=1

= Pn

i=1

k¯ xkp¯,1 =

i=1

pi |xi | =

n X

1

k=1

n X

n X

a2k

n X i=1

a2i

bi − c Pn 2 k=1 ak ai

|ai | |bi − cai | ,

i=1

n X bi = Pn1 Pn |ai bi | 2 2 k=1 ak ai k=1 ak i=1 a2i

and 1 k¯ x − x¯µ,p kp¯,1 = Pn 2 ( k=1 a2k )

n X i=1

n X a a |ai | ak k i . bk bi k=1

Utilising the inequality (4.120) we deduce (4.119).

4.16

Another Counterpart via a Gr¨ uss Type Result

The following Gr¨ uss type inequality has been obtained in [21]. ¯ = (b1 , . . . , bn ) be two sequences of real Lemma 195 Let ¯ a = (a1 , . . . , an ), b numbers and assume that there are γ, Γ ∈ R such that −∞ < γ ≤ ai ≤ Γ < ∞ for each i ∈ {1, . . . , n} .

(4.121)

Then for any p ¯ = (p1 , . . . , pn ) a nonnegative sequence with the property that P n i=1 pi = 1, one has the inequality n n n n n 1 X X X X X p i ai b i − p i ai pi bi ≤ (Γ − γ) p i bi − pk bk . (4.122) 2 i=1 i=1 i=1 i=1 k=1

148

CHAPTER 4. COUNTERPART INEQUALITIES

The constant constant.

1 2

is sharp in the sense that it cannot be replaced by a smaller

Proof. We will give here a simpler direct proof based on Sonin’s identity. A simple calculation shows that: n X

p i ai b i −

i=1

n X

p i ai

i=1

n X

p i bi

i=1

=

n X

 pi

i=1

γ+Γ ai − 2

 bi −

n X

! p k bk

. (4.123)

k=1

By (4.121) we have Γ−γ γ + Γ ≤ ai − for all i ∈ {1, . . . , n} 2 2 and thus, by (4.123), on taking the modulus, we get n n n n n X X X X X γ + Γ bi − p i ai b i − pi ai p i bi ≤ pi ai − p b k k 2 i=1 i=1 i=1 i=1 k=1 n n X X 1 ≤ (Γ − γ) p i bi − p k bk . 2 i=1

k=1

To prove the sharpness of the constant 21 , let us assume that (4.122) holds with a constant c > 0, i.e., n n n n n X X X X X pi ai bi − p i ai pi bi ≤ c (Γ − γ) p i bi − p k bk . i=1

i=1

i=1

i=1

k=1

provided ai satisfies (4.121). If we choose n = 2 in (4.124) and take into account that 2 X i=1

p i ai b i −

2 X i=1

p i ai

2 X i=1

pi bi = p1 p2 (a1 − a2 ) (b1 − b2 )

(4.124)

¨ TYPE RESULT 4.16. ANOTHER COUNTERPART VIA A GRUSS

149

provided p1 + p2 = 1, p1 , p2 ∈ [0, 1] , and since 2 2 X X p i bi − pk bk = p1 |(p1 + p2 ) b1 − p1 b1 − p2 b2 | i=1

k=1

+ p2 |(p1 + p2 ) b2 − p1 b1 − p2 b2 | = 2p1 p2 |b1 − b2 | we deduce by (4.124) p1 p2 |a1 − a2 | |b1 − b2 | ≤ 2c (Γ − γ) |b1 − b2 | p1 p2 .

(4.125)

If we assume that p1 , p2 6= 0, b1 6= b2 and a1 = Γ, a2 = γ, then by (4.125) we deduce c ≥ 12 , which proves the sharpness of the constant 12 . The following corollary is a natural consequence of the above lemma. Corollary 196 Assume that ¯ a = (a1 , . . . , an ) satisfies the assumption (4.121) and p ¯ is a probability sequence. Then !2 n n n n X X X X 1 2 0≤ p i ai − pi ai ≤ (Γ − γ) p i ai − p k ak . (4.126) 2 i=1

The constant

i=1

1 2

i=1

k=1

is best possible in the sense mentioned above.

The following counterpart of the (CBS) −inequality may be stated. Theorem 197 Assume that x ¯ = (x1 , . . . , xn ) and y ¯ = (y1 , . . . , yn ) are sequences of real numbers with yi 6= 0 (i = 1, . . . , n) . If there exists the real numbers m, M such that m≤

xi ≤ M for each i ∈ {1, . . . , n} , yi

(4.127)

then we have the inequality 0≤

n X i=1

x2i

n X

yi2 −

i=1

1 ≤ (M − m) 2

n X

!2 xi yi

i=1 n X i=1

n X x y |yi | yk · i i . xk yk k=1

(4.128)

150

CHAPTER 4. COUNTERPART INEQUALITIES

Proof. If we choose pi = Γ = M in (4.126), we deduce Pn 2 x Pni=1 i2 − k=1 yk

k=1

k=1

n X

1 Pn

y2 Pn i

yk2

yk2

, ai =

xi yi

for i = 1, . . . , n and γ = m,

!2 xi yi

i=1

n X 1 1 1 2 xi P P ≤ (M − m) n yi − n xk yk 2 2 yi 2 k=1 yk i=1 k=1 yk k=1 n n n X X X 1 1 2 = (M − m) Pn |yi | xi yk − yi xk yk 2 2 2 ( k=1 yk ) i=1 k=1 k=1 n n X X xi yi 1 1 . = (M − m) Pn |yi | yk · 2 x k y k 2 ( k=1 yk2 ) i=1 k=1 n X

giving the desired inequality (4.128).

Bibliography [1] R. FRUCHT, Sobre algunas disigualdades: Observaci´on relative a la soluci´on del Problema No. 21, indicada par el Ing. Ernesto M. Saleme (in Spanish), Math. Notae, 3 (1943), 41-46. [2] G.S. WATSON, Serial Correlation in Regression Analysis, Ph.D. Thesis, Dept. of Experimental Statistics, North Carolina State College, Raleigh; Univ. of North Carolina, Mimograph Ser., No. 49, 1951. [3] G.S. WATSON, Serial correlation in regression analysis, I, Biometrika, 42 (1955), 327-341. [4] G.S. WATSON, A method for discovering Kantorovich-type inequalities and a probabilistic interpretation, Linear Algebra Appl., 97 (1987), 211217. [5] G.S. WATSON, G. ALPARGU and G.P.H. STYAN, Some comments on six inequalities associated with the inefficiency of ordinary least squares with one regressor, Linear Algebra and its Appl., 264 (1997), 13-54. ´ ¨ Aufgaben und Lehrs¨ [6] G. POLYA and G. SZEGO, utze ans der Analysis, Band I: Reihen, Integralrechnung, Funktiontheorie (in German), 4th Ed., Springer Verlag, Berlin, 1970 (Original version: Julius Springer, Berlin, 1925). ´ ¨ Problems and Theorems in Analysis, Vol[7] G. POLYA and G. SZEGO, ume 1: Series, Integral Calculus, Theory of Functions (in English), translated from german by D. Aeppli, corrected printing of the revised translation of the fourth German edition, Springer Verlag, New York, 1972. 151

152

BIBLIOGRAPHY

[8] W. GREUB and W. RHEINBOLDT, On a generalisation of an inequality of L.V. Kantorovich, Proc. Amer. Math. Soc., 10 (1959), 407-415. [9] S.S. DRAGOMIR, A generalisation of Cassel’s and Greub-Reinboldt’s inequalities in inner product spaces, submitted. [10] S.S. DRAGOMIR, A counterpart of Schwartz’s inequality in inner product spaces, submitted. [11] S.S. DRAGOMIR and G.L. BOOTH, Gr¨ uss-Lupa¸s type inequality and its applications for the estimation of p-moments of guessing mappings, Math. Comm., 5 (2000), 117-126. [12] S.S. DRAGOMIR, Some counterpart inequalities in terms of forward difference for (CBS) −inequality, submitted. [13] S.S. DRAGOMIR, A Gr¨ uss type inequality for sequences of vectors in normed linear spaces and applications, submitted. [14] S.S. DRAGOMIR, Another Gr¨ uss type inequality for sequences of vectors in normed linear spaces and applications, J. Comp. Anal. & Applic., 4 (2) (2002), 155-172. [15] D. ANDRICA and C. BADEA, Gr¨ uss’ inequality for positive linear functionals, Periodica Mat. Hungarica, 19(2) (1988), 155-167. [16] A.W. ROBERTS and D.E. VARBERG, Convex Functions, Academic Press, New York, 1973. [17] A. LUPAS¸, A remark on the Schweitzer and Kantorovich inequalities, Publ. Elek. Fak. Univ. Beograde, Ser. Mat. i Fiz., 383 (1972), 13-14. [18] P. SCHWEITZER, An inequality about the arithmetic mean (Hungarian), Math. Phys. Lapok (Budapest), 23 (1914), 257-261. [19] J.B. DIAZ and F.T METCALF, Stronger forms of a class of inequalities of G. P´olya – G. Szeg¨o and L.V. Kantorovich, Bull. Amer. Math. Soc., 69 (1963), 415-418. [20] P. CERONE and S.S. DRAGOMIR, New inequalities for ˇ the Cebyˇ sev functional involving two n−tuples of real numbers and applications, RGMIA Res. Rep. Coll., 5(4) (2002), http://rgmia.vu.edu.au/v5n4.html.

BIBLIOGRAPHY

153

[21] P. CERONE and S.S. DRAGOMIR, A refinement of the Gr¨ uss inequality and applications, RGMIA Res. Rep. Coll., 5(2) (2002), Article 2. [ONLINE: http://rgmia.vu.edu.au/v5n2.html] ˇ [22] S.S. DRAGOMIR, New inequalities for the weighted Cebyˇ sev functionals and applications for the (CBS) −inequality, in preparation. [23] O. SHISHA and B. MOND, Bounds on differences of means, Inequalities, Academic Press Inc., New York, 1967, 293-308. [24] D. ZAGIER, A converse to Cuachy’s inequality, Amer. Math. Month., 102(10) (1995), 919-920. [25] D. ZAGIER, An inequality converse to that of Cauchy (Dutch), Indag. Math., 39(4) (1997), 349-351. ˇ ´ The inequalities of Zagier and Cebyˇ ˇ [26] J. PECARI C, sev, Arch. Math., (Basel), 64(5) (1995), 415-417. [27] H. ALZER, On a converse Cauchy inequality of D. Zagier, Arch. Math., (Basel), 58(2) (1995), 157-159.

154

BIBLIOGRAPHY

Chapter 5 Related Inequalities 5.1

Ostrowski’s Inequality for Real Sequences

In 1951, A.M. Ostrowski [2, p. 289] gave the following result related to the (CBS) −inequality for real sequences (see also [1, p. 92]). ¯ = (b1 , . . . , bn ) be two non-proportional Theorem 198 Let ¯ a = (a1 , . . . , an ) and b sequences of real numbers. Let x ¯ = (x1 , . . . , xn ) be a sequence of real numbers such that n n X X ai xi = 0 and bi xi = 1. (5.1) i=1

Then

n X

x2i

i=1

Pn

≥ Pn

i=1

i=1

2 i=1 ai Pn 2 i=1 bi − ( i=1

Pn 2

ai

2

ai b i )

(5.2)

with equality if and only if bk

xk = Pn

Pn

2 i=1 ai

i=1

a2i − ak

Pn

2 i=1 bi

Pn

ai b i Pi=1 2 n − ( i=1 ai bi )

(5.3)

for any k ∈ {1, . . . , n} . Proof. We shall follow the proof in [1, p. 93 – p. 94]. Let n n n X X X 2 2 bi , C = ai b i A= ai , B = i=1

i=1

155

i=1

(5.4)

156

CHAPTER 5. RELATED INEQUALITIES

and yi =

Abi − Cai for any i ∈ {1, . . . , n} . AB − C 2

(5.5)

It is easy to see that the sequence y ¯ = (y1 , . . . , yn ) as defined by (5.5) satisfies (5.1). Any sequence x ¯ = (x1 , . . . , xn ) that satisfies (5.1) fulfills the equality n X

xi yi =

n X

i=1

xi ·

i=1

(Abi − Cai ) A = ; 2 AB − C AB − C 2

so, in particular n X

A . AB − C 2

yi2 =

i=1

Any sequence x ¯ = (x1 , . . . , xn ) that satisfies (5.1) therefore satisfies n X

x2i

−

i=1

n X

yi2

=

i=1

n X

(xi − yi )2 ≥ 0,

(5.6)

i=1

and thus n X i=1

x2i ≥

n X i=1

yi2 =

A AB − C 2

and the inequality (5.2) is proved. From (5.6) it follows that equality holds in (5.1) iff xi = yi for each i ∈ {1, . . . , n}, and the theorem is completely proved.

5.2

Ostrowski’s Inequality for Complex Sequences

The following result that points out a natural generalisation of Ostrowski’s inequality for complex numbers holds [3]. ¯ = (b1 , . . . , bn ) and x Theorem 199 Let ¯ a = (a1 , . . . , an ) , b ¯ = (x1 , . . . , xn )
¯ ¯ be sequences of complex numbers. If ¯ a and b, where b = ¯b1 , . . . , ¯bn , are

5.2. OSTROWSKI’S INEQUALITY FOR COMPLEX SEQUENCES 157 not proportional and n X

xi a ¯i = 0;

(5.7)

i=1 n X xi¯bi = 1,

(5.8)

i=1

then one has the inequality n X i=1

Pn

2

|xi | ≥ Pn

i=1

i=1

|ai |2

Pn

i=1

|ai |2

P 2 |bi |2 − ni=1 ai¯bi

(5.9)

with equality iff # Pn b a ¯ k k xi = µ bi − Pnk=1 2 · ai , i ∈ {1, . . . , n} k=1 |ak | "

(5.10)

and µ ∈ C with Pn |µ| = Pn

k=1

2 k=1 |ak |

|ak |2

Pn 2 . 2 ¯ |b | − a b k=1 k k=1 k k

Pn

Proof. Recall the (CBS) −inequality for complex sequences 2 n n n X X X 2 2 uk v¯k |uk | |vk | ≥ k=1

k=1

(5.11)

(5.12)

k=1

with equality iff there is a complex number α ∈ C such that uk = αvk , k = 1, . . . , n. If we apply (5.12) for Pn zi c¯i uk = zk − Pni=1 2 · ck , |ci | Pi=1 n di c¯i ¯ ¯ vk = dk − Pni=1 2 · ck , where ¯ c 6= 0 and ¯ c, d, z ∈ Cn , |c | i=1 i

(5.13)

158

CHAPTER 5. RELATED INEQUALITIES

we have 2 n 2 Pn Pn n X X d c ¯ z c ¯ i i i i i=1 i=1 · ck · ck dk − Pn zk − Pn 2 2 i=1 |ci | i=1 |ci | k=1 k=1 ! ! 2 Pn Pn n X z c ¯ d c ¯ i i i i ≥ zk − Pni=1 2 · ck dk − Pni=1 2 · ck (5.14) i=1 |ci | i=1 |ci | k=1

with equality iff there is a β ∈ C such that ! Pn Pn z c ¯ d c ¯ i i i i zk = Pni=1 2 · ck + β dk − Pni=1 2 · ck . i=1 |ci | i=1 |ci |

(5.15)

Since a simple calculation shows that 2 P Pn Pn n 2 n 2 Pn 2 X z c ¯ ¯k | i i k=1 |zk | k=1 |ck | − | k=1 zk c i=1 , zk − Pn Pn 2 · ck = 2
2 |c | i=1 |ci | k k=1 k=1 2 P P P Pn n 2 n 2 n 2 n X ¯k | ¯i i=1 di c k=1 |dk | k=1 |ck | − | k=1 dk c dk − Pn
2 Pn 2 · ck = |ck |2 i=1 |ci | k=1

k=1

and n X k=1

! ! Pn Pn z c ¯ d c ¯ i i i i zk − Pni=1 2 · ck dk − Pni=1 2 · ck i=1 |ci | i=1 |ci | Pn ¯ Pn |ck |2 − Pn zk c¯k · Pn ck d¯k k=1 zk dk · k=1 k=1 k=1 = Pn 2
2 i=1 |ci |

then by (5.12) we deduce  2   2  n n n n n n X X X X X X  2 2 2 2  |zk | |ck | − zk c¯k |dk | |ck | − dk c¯k  k=1 k=1 k=1 k=1 k=1 k=1 2 n n n n X X X X ≥ zk d¯k · |ck |2 − zk c¯k ck d¯k (5.16) k=1

k=1

k=1

with equality iff there is a β ∈ C such that (5.15) holds.

k=1

5.3. ANOTHER OSTROWSKI’S INEQUALITY

159

¯ satisfy (5.7) and (5.8), then by (5.16) and (5.15) for the choices If ¯ a, x ¯, b ¯ = b, ¯ we deduce (5.9) with equality iff there is a µ ∈ C ¯ z=x ¯, ¯ c=¯ a and d such that ! Pn ¯ a b i i xk = µ bk − Pni=1 2 · ak , i=1 |ai | and, by (5.8), n ! Pn X ¯bi a i bk − Pni=1 2 · ak · ¯bk = 1. µ i=1 |ai |

(5.17)

k=1

Since (5.17) is clearly equivalent to (5.15), the theorem is completely proved.

5.3

Another Ostrowski’s Inequality

In his book from 1951, [2, p. 130], A.M. Ostrowski proved the following inequality as well (see also [1, p. 94]). ¯ x Theorem 200 Let ¯ a, b, ¯ be sequences of real numbers so that ¯ a 6= 0 and n X

x2k = 1

(5.18)

ak xk = 0.

(5.19)

k=1 n X k=1

Then Pn

k=1

a2k

Pn

Pn

b2 − ( k=1 Pnn 2 k=1 ak

k=1

2

ak b k )

≥

n X

!2 bk xk

.

(5.20)

k=1

¯ are non-proportional, then equality holds in (5.20) iff If ¯ a and b P P bk ni=1 a2i − ak ni=1 ai bi xk = q · hP i 12 , 1 P P Pn 2 n n n 2 2 ( k=1 a2k ) 2 i=1 ai i=1 bi − ( i=1 ai bi ) k ∈ {1, . . . , n} , q ∈ {−1, 1} . (5.21) We may extend this result for sequences of complex numbers as follows [4].

160

CHAPTER 5. RELATED INEQUALITIES

¯ x Theorem 201 Let ¯ a, b, ¯ be sequences of complex numbers so that ¯ a 6= 0, ¯ ¯ a, b are not proportional, and n X k=1 n X

|xk |2 = 1

(5.22)

xk a ¯k = 0.

(5.23)

k=1

Then Pn

2 k=1 |ak |

2 Pn 2 n 2 ¯ X |b | − a b k=1 k k=1 k k ¯ ≥ x k bk . Pn 2 k=1 |ak | k=1

Pn

The equality holds in (5.24) iff ! Pn b a ¯ i i xk = β bk − Pni=1 2 · ak , i=1 |ai |

k ∈ {1, . . . , n} ;

(5.24)

(5.25)

where β ∈ C is such that Pn |β| =  Pn

k=1

2 k=1 |ak |

2

|ak |


21

Pn 2  12 ¯ |b | − a b k=1 k k=1 k k

Pn

.

(5.26)

2

Proof. In Section 5.2, we proved the following inequality:  2  n n n X X X  |zk |2 |ck |2 − zk c¯k  k=1 k=1 k=1  2  n n n X X X × |dk |2 |ck |2 − dk c¯k  k=1 k=1 k=1 2 n n n n X X X X 2 ¯ ¯ ≥ zk dk · |ck | − zk c¯k ck dk (5.27) k=1

k=1

k=1

k=1

¯ sequences of complex numbers, with equality iff there is a for any ¯ z, ¯ c, d β ∈ C such that ! Pn Pn d c ¯ z c ¯ i i i i (5.28) zk = Pni=1 2 · ck + β dk − Pni=1 2 · ck i=1 |ci | i=1 |ci |

5.4. FAN AND TODD INEQUALITIES

161

for each k ∈ {1, . . . , n} . ¯=b ¯ and take into consideration If in (5.27) we choose ¯ z=x ¯, ¯ c=¯ a and d that (5.22) and (5.23) hold, then we get  2  n n n n n X X X X X 2 2 2 2 bk a ¯k  |bk | |ak | − |ak | |xk | k=1 k=1 k=1 k=1 k=1 2 !2 n n X X ≥ xk¯bk |ak |2 k=1

which is clearly equivalent to (5.24). By (5.28) the equality holds in (5.24) iff ! Pn b a ¯ i i xk = β bk − Pni=1 2 · ak , i=1 |ai |

k=1

k ∈ {1, . . . , n} .

Since x ¯ should satisfy (5.22), we get 2 Pn n X b a ¯ i i i=1 |xk | = |β| 1= · ak bk − Pn 2 i=1 |ai | k=1 k=1 # " n P 2 X | nk=1 bk a ¯k | 2 2 = |β| |bk | − Pn 2 k=1 |ak | k=1 n X

2

2

from where we deduce that β satisfies (5.26).

5.4

Fan and Todd Inequalities

In 1955, K. Fan and J. Todd [5] proved the following inequality (see also [1, p. 94]). ¯ = (b1 , . . . , bn ) be two sequences of Theorem 202 Let ¯ a = (a1 , . . . , an ) and b real numbers such that ai bj 6= aj bi for i 6= j. Then Pn 2 ai Pn 2 Pn i=12 P 2 ( i=1 ai ) ( i=1 bi ) − ( ni=1 ai bi )  2  X n n aj n X  ≤   . (5.29) 2 aj b i − ai b j j=1 i=1 j6=i

162

CHAPTER 5. RELATED INEQUALITIES

Proof. We shall follow the proof in [1, p. 94 – p. 95]. Define  −1 X aj n (1 ≤ i ≤ n) . xi := 2 aj b i − ai b j j6=i

The terms in the sum on the right-hand side     n n n −1 X X ai aj n  X x i ai =   2 a b − a b j i i j j=1 i=1

i=1

j6=i

can be grouped in pairs of the form 

n 2

−1 

aj ai ai aj + aj b i − ai b j ai b j − aj b i

 (i 6= j)

and the sum of each such pair vanishes. Hence, we deduce n X

ai xi = 0 and

i=1

n X

bi xi = 1.

i=1

Applying Ostrowski’s inequality (see Section 5.1) we deduce the desired result (5.29). A weighted version of the result is also due to K. Fan and J. Todd [5] (see also [1, p. 95]). We may state the result as follows. Theorem 203 Let pij (i, j ∈ {1, . . . , n} , i 6= j) be real numbers such that pij = pji , for any i, j ∈ {1, . . . , n} with i 6= j. (5.30) P Denote P := 1≤i<j≤n pij and assume that P 6= 0. Then for any two se¯ = (b1 , . . . , bn ) satisfying quences of real numbers ¯ a = (a1 , . . . , an ) and b ai bj 6= aj bi (i 6= j) , we have Pn

2 i=1 ai Pn 2 Pn 2 Pn i=1 ai i=1 bi − ( i=1

2

ai b i )

≤

1 P2

n X i=1

  

n X j=1 j6=i

2 pij aj   . aj b i − ai b j

(5.31)

5.5. SOME RESULTS FOR ASYNCHRONOUS SEQUENCES

5.5

163

Some Results for Asynchronous Sequences

If S (R) is the linear space of real sequences, S+ (R) is the subset of nonnegative sequences and Pf (N) denotes the set of finite parts of N, then for the functional T : Pf (N) × S+ (R) × S 2 (R) → R, ! 21 X X X pi a2i pi b2i − p i ai b i (5.32) T (I, p, a, b) := i∈I

i∈I

i∈I

we may state the following result [6, Theorem 3]. Theorem 204 If |¯ a| = (|ai |)i∈N and ¯ b = (|bi |)i∈N are asynchronous, i.e., (|ai | − |aj |) (|bi | − |bj |) ≤ 0 for all i, j ∈ N, then P P X i∈I pi |ai | i∈I pi |bi | P T (I, p, a, b) ≥ − pi |ai bi | ≥ 0. (5.33) i∈I pi i∈I Proof. We shall follow the proof in [6]. Consider the inequalities ! 12 X X X pi pi a2i ≥ pi |ai | i∈I

i∈I

X

X

i∈I

and

! 12 i∈I

pi

pi b2i

≥

i∈I

X

pi |bi |

i∈I

that by multiplication give ! 12 X i∈I

pi a2i

X i∈I

pi b2i

P ≥

i∈I

pi |ai | P

P

i∈I

i∈I

pi

pi |bi |

.

ˇ Now, by the definition of T and by Cebyˇ sev’s inequality for asynchronous sequences, we have P P X p |a | p |b | i i i i i∈I − pi ai bi T (I, p, a, b) ≥ i∈I P p i∈I i i∈I P P pi |ai | i∈I pi |bi | X ≥ i∈I P − pi |ai | |bi | i∈I pi i∈I ≥0

164

CHAPTER 5. RELATED INEQUALITIES

and the theorem is proved. The following result also holds [6, Theorem 4]. Theorem 205 If |¯ a| and ¯ b are synchronous, i.e., (|ai | − |aj |) (|bi | − |bj |) ≥ 0 for all i, j ∈ N, then one has the inequality 0 ≤ T (I, p, a, b) ≤ T (I, p, ab, 1) ,

(5.34)

where 1 = (ei )i∈N , ei = 1, i ∈ N. ˇ Proof. We have, by Cebyˇ sev’s inequality for the synchronous sequences 2 2 2 ¯ = (bi ) , that ¯ a = (ai )i∈N and b i∈N ! 21 X X X 2 2 T (I, p, a, b) = p i ai p i bi − p i ai b i i∈I i∈I i∈I 1 !2 X X X 2 2 ≤ pi ai bi pi − p i ai b i 2

i∈I

i∈I

i∈I

= T (I, p, ab, 1) and the theorem is proved.

5.6

An Inequality via A − G − H Mean Inequality

The following result holds [6, Theorem 5]. ¯ be sequences of positive real numbers. Define Theorem 206 Let ¯ a and b 2 ai b2i ∆i = P (5.35) P 2 2 a b i∈I i i∈I i where i ∈ I and I is a finite part of N. Then one has the inequality "   # P 21P 2
2 P Y ai ∆i i∈I ai i∈I bi a b i i P i∈I2 P ≥ 2 bi i∈I ai i∈I bi i∈I P P 2 b2 i∈I ai ≥ P a3 Pi∈I b3i . i

i

i∈I bi

i∈I ai

(5.36)

5.6. AN INEQUALITY VIA A − G − H MEAN INEQUALITY

165

The equality holds in all the inequalities from (5.36) iff there exists a positive number k > 0 such that ai = kbi for all i ∈ I. Proof. We shall follow the proof in [6]. We will use the AGH−inequality 1 X pi xi ≥ PI i∈I

! P1 Y

I

xpi i

≥P

PI

pi i∈I xi

i∈I

,

(5.37)

P where pi > 0, xi ≥ 0 for all i ∈ I, where PI := i∈I pi > 0. We remark that the equality holds in (5.37) iff xi = xj for each i, j ∈ I. Choosing pi = a2i and xi = abii (i ∈ I) in (5.37), then we get a2 P P 2 Y  bi  Pi∈Ii a2i a b i i i∈I i∈I ai P ≥ ≥ P a3i 2 ai i∈I ai i∈I

(5.38)

i∈I bi

and by pi = b2i and xi =

ai , bi

we also have

P P   P b2i 2 2 Y a b ai i∈I bi i∈I bi i∈I i i P ≥ P b3 . ≥ 2 i bi i∈I bi i∈I

(5.39)

i∈I ai

If we multiply (5.38) with (5.39) we easily deduce the desired inequality (5.36). The case of equality follows by the same case in the arithmetic mean – geometric mean – harmonic mean inequality. We omit the details. The following corollary holds [6, Corollary 5.1]. ¯ as above, one has the inequality Corollary 207 With ¯ a and b P

b3i i∈I ai
2 P i∈I ai bi

a3i i∈I bi



P

 12  ≥

P

2 i∈I bi
2 . P a b i i i∈I

i∈I

a2i

P

(5.40)

The equality holds in (5.40) iff there is a k > 0 such that ai = kbi , i ∈ {1, . . . , n} .

166

5.7

CHAPTER 5. RELATED INEQUALITIES

A Related Result via Jensen’s Inequality for Power Functions

The following result also holds [6, Theorem 6]. ¯ be sequences of positive real numbers and p ≥ 1. Theorem 208 Let ¯ a and b If I is a finite part of N, then one has the inequality "P #1
2 2−p p P p 2−p p a b a b a b i i i i∈I i P i∈I2 P P Pi∈I 2i i ≤ . 2 2 i∈I ai i∈I ai i∈I bi i∈I bi P

(5.41)

The equality holds in (5.41) if and only if there exists a k > 0 such that ai = kbi for all i ∈ I. If p ∈ (0, 1) , the inequality in (5.41) reverses. Proof. We shall follow the proof in [6]. By Jensen’s inequality for the convex mapping f : R+ → R, f (x) = xp , p ≥ 1 one has

P

i∈I pi xi PI

p

P ≤

i∈I

pi xpi

PI

,

(5.42)

P where PI := i∈I pi , pi > 0, xi ≥ 0, i ∈ I. The equality holds in (5.42) iff xi = xj for all i, j ∈ I. Now, choosing in (5.42) pi = a2i , xi = abii , we get P ai b i Pi∈I 2 ≤ i∈I ai and for pi = b2i , xi =

ai , bi

! p1 P 2−p p a b i i∈I i P 2 i∈I ai

(5.43)

the inequality (5.42) also gives

P ai b i Pi∈I 2 ≤ i∈I bi

! p1 P p 2−p a b i∈I i i P . 2 i∈I bi

(5.44)

By multiplying the inequalities (5.43) and (5.44), we deduce the desired result from (5.42).

5.8. INEQUALITIES DERIVED FROM THE DOUBLE SUMS CASE 167 The case of equality follows by the fact that in (5.42) the equality holds iff (xi )i∈I is constant. If p ∈ (0, 1) , then a reverse inequality holds in (5.42) giving the corresponding result in (5.41). Remark 209 If p = 2, then (5.41) becomes the (CBS) −inequality.

5.8

Inequalities Derived from the Double Sums Case

Let A = (aij )i,j=1,n and B = (bij )i,j=1,n be two matrices of real numbers. The following inequality is known as the (CBS) −inequality for double sums !2 n n n X X X aij bij ≤ a2ij b2ij (5.45) i,j=1

i,j=1

i,j=1

with equality iff there is a real number r such that aij = rbij for any i, j ∈ {1, . . . , n} . The following inequality holds [7, Theorem 5.2]. ¯ = (b1 , . . . , bn ) be sequences of real Theorem 210 Let ¯ a = (a1 , . . . , an ) and b numbers. Then !2 !2 n n n X X X a + b − 2n a b k k k k k=1 k=1 k=1 ≤n

n X

a2k

+

b2k




−2

k=1

n X

ak

k=1

n X

bk . (5.46)

k=1

Proof. We shall follow the proof from [7]. Applying (5.45) for aij = ai − bj , bij = bi − aj and taking into account that !2 !2 n n n n X X X X (ai − bj ) (bi − aj ) = 2n ak b k − ak − bk i,j=1 n X i,j=1

(ai − bj )2 = n

k=1 n X k=1

k=1


a2k + b2k − 2

k=1 n X k=1

ak

n X k=1

bk

168

CHAPTER 5. RELATED INEQUALITIES

and

n X

n X

2

(bi − aj ) = n

i,j=1

a2k

+

b2k




−2

n X

k=1

ak

k=1

n X

bk ,

k=1

we may deduce the desired inequality (5.46). The following result also holds [7, Theorem 5.3]. ¯ = (b1 , . . . , bn ) , ¯ Theorem 211 Let ¯ a = (a1 , . . . , an ), b c = (c1 , . . . , cn ) and ¯ d = (d1 , . . . , dn ) be sequences of real numbers. Then one has the inequality: 

 Pn

ai c i

Pn

i=1 bi ci

Pn

i=1

det  Pn

i=1

ai di

2 

i=1 bi di

 Pn

Pn

≤ det  P n

Pn

2 i=1 ai

i=1

i=1 ai bi

ai b i 

 

b2i Pni=1 2 i=1 ci

Pn

i=1 ci di

× det  P n

Pn

2 i=1 di

i=1 ci di

  . (5.47)

Proof. We shall follow the proof in [7]. Applying (5.45) for aij = ai bj − aj bi , bij = ci dj − cj di and using CauchyBinet’s identity [1, p. 85] n 1X (ai bj − aj bi ) (ci dj − cj di ) 2 i,j=1

=

n X

ai c i

i=1

n X i=1

bi di −

n X

! ai di

i=1

n X

! bi ci

(5.48)

i=1

and Lagrange’s identity [1, p. 84] n n n X X 1X 2 2 (ai bj − aj bi ) = ai b2i − 2 i,j=1 i=1 i=1

we deduce the desired result (5.47).

n X i=1

!2 ai b i

,

(5.49)

5.9. A FUNCTIONAL GENERALISATION FOR DOUBLE SUMS

5.9

169

A Functional Generalisation for Double Sums

The following result holds [7, Theorem 5.5]. Theorem 212 Let A be a subset of real numbers R, f : A → R and ¯ a= ¯ (a1 , . . . , an ), b = (b1 , . . . , bn ) sequences of real numbers with the property that (i) ak bi , a2i , b2k ∈ A for any i, k ∈ {1, . . . , n} ; (ii) f (a2k ) , f (b2k ) ≥ 0 for any k ∈ {1, . . . , n} ; (iii) f 2 (ak bi ) ≤ f (a2k ) f (b2i ) for any i, k ∈ {1, . . . , n} . Then one has the inequality #2 " n n n X X
X
2 2 f ak f b2k . f (ak bi ) ≤ n k=1

k,i=1

(5.50)

k=1

Proof. We will follow the proof in [7]. Using the assumption (iii) and the (CBS) −inequality for double sums, we have n n n X X X 

 1 |f (ak bi )| ≤ f a2k f b2i 2 (5.51) f (ak bi ) ≤ k,i=1 k,i=1 k,i=1  !2 !2  12 n n  X X
 1  
 1 ≤ f a2k 2 f b2i 2   k,i=1

k,i=1

"

n X

=

f a2k

n
X

k,i=1

" =n

n X k=1

# 12 f b2i




k,i=1

# 12 "
2

f ak

n X

# 21
2

f bk

k=1

which is clearly equivalent to (5.50). The following corollary is a natural consequence of the above theorem [7, Corollary 5.6].

170

CHAPTER 5. RELATED INEQUALITIES

Corollary 213 Let A, f and ¯ a be as above. If (i) ak ai ∈ A for any i, k ∈ {1, . . . , n} ; (ii) f (a2k ) ≥ 0 for any k ∈ {1, . . . , n} ; (iii) f 2 (ak ai ) ≤ f (a2k ) f (a2i ) for any i, k ∈ {1, . . . , n} , Then one has the inequality n n X X
f (ak ai ) ≤ n f a2k . k,i=1

(5.52)

k=1

The following particular inequalities also hold [7, p. 23]. 1. If ϕ : N → N is Euler’s indicator and s (n) denotes the sum of all relatively prime numbers including and less than n, then for any ¯ a= ¯ (a1 , . . . , an ), b = (b1 , . . . , bn ) sequences of natural numbers, one has the inequalities " n #2 n n X X
X
2 2 ϕ (ak bi ) ≤ n ϕ ak ϕ b2k ; (5.53) k,i=1 n X

k=1 n X

ϕ (ak ai ) ≤ n

k,i=1

"

n X


ϕ a2k ;

(5.54)

k=1

#2 s (ak bi )

k,i=1 n X

k=1

≤n

2

n X

s

a2k

n
X

k=1

s (ak ai ) ≤ n

k,i=1

n X


s b2k ;

(5.55)

k=1


s a2k .

(5.56)

k=1

¯ = (b1 , . . . , bn ) are sequences of real 2. If a > 1 and ¯ a = (a1 , . . . , an ), b numbers, then " n #2 n n X X

X 2 2 expa (ak bi ) ≤ n expa b2k ; (5.57) expa ak k,i=1 n X k,i=1

expa (ak ai ) ≤ n

k=1 n X


expa a2k ;

k=1

k=1

(5.58)

5.10. A (CBS) −TYPE RESULT FOR LIPSCHITZIAN FUNCTIONS 171 ¯ are sequences of real numbers such that ak , bk ∈ (−1, 1) 3. If ¯ a and b (k ∈ {1, . . . , n}) , then one has the inequalities: "

n X

1 (1 − ak bi )m k,i=1

#2 ≤n

2

n X

k=1 n X

n X

n X 1 1 m, 2 m (1 − ak ) k=1 (1 − b2k )

1 1 m, m ≤ n (1 − ak ai ) (1 − a2k ) k,i=1 k=1

(5.59) (5.60)

where m > 0.

5.10

A (CBS) −Type Result for Lipschitzian Functions

The following result was obtained in [8, Theorem]. Theorem 214 Let f : I ⊆ R → R be a Lipschitzian function with the constant M, i.e., it satisfies the condition |f (x) − f (y)| ≤ M |x − y| for any x, y ∈ I.

(5.61)

¯ = (b1 , . . . , bn ) are sequences of real numbers with ai bj ∈ If ¯ a = (a1 , . . . , an ), b I for any i, j ∈ {1, . . . , n} , then n n X X 0≤ f (ai bj ) |f (ai bj )| − |f (aj bi )| f (ai bj ) (5.62) i,j=1

≤

n X

i,j=1

f 2 (aj bi ) −

i,j=1

n X

f (ai bj ) f (aj bi )

i,j=1

 n n X X ≤ M2  a2i b2i − i=1

i=1

n X

!2  ai b i

.

i=1

Proof. We shall follow the proof in [8]. Since f is Lipschitzian with the constant M, we have 0 ≤ ||f (ai bj )| − |f (aj bi )|| ≤ |f (ai bj ) − f (aj bi )| ≤ M |ai bj − aj bi |

(5.63)

172

CHAPTER 5. RELATED INEQUALITIES

for any i, j ∈ {1, . . . , n} , giving 0 ≤ |(|f (ai bj )| − |f (aj bi )|) (f (ai bj ) − f (aj bi ))|

(5.64)

≤ (f (ai bj ) − f (aj bi ))2 ≤ M 2 (ai bj − aj bi )2 for any i, j ∈ {1, . . . , n} . The inequality (5.64) is obviously equivalent to |f (ai bj )| f (ai bj ) + |f (aj bi )| f (aj bi )

(5.65)

− |f (ai bj )| f (ai bj ) − |f (aj bi )| f (aj bi ) ≤ f 2 (ai bj ) − 2f (ai bj ) f (aj bi ) + f 2 (aj bi )
≤ M 2 a2i b2j − 2ai bi aj bj + a2j b2i for any i, j ∈ {1, . . . , n} . Summing over i and j from 1 to n in (5.65) and taking into account that: n X i,j=1 n X

n X

|f (ai bj )| f (ai bj ) = |f (ai bj )| f (aj bi ) =

i,j=1 n X

f 2 (ai bj ) =

i,j=1

|f (aj bi )| f (ai bj ) ,

i,j=1 n X i,j=1 n X

|f (aj bi )| f (ai bj ) , f 2 (aj bi ) ,

i,j=1

we deduce the desired inequality. The following particular inequalities hold [8, p. 27 – p. 28]. 1. Let x ¯ = (x1 , . . . , xn ) , y ¯ = (y1 , . . . , yn ) be sequences of real numbers such that 0 ≤ |xi | ≤ M1 , 0 ≤ |yi | ≤ M2 , i ∈ {1, . . . , n} . Then for any r ≥ 1 one has !2 n n n X X X 0≤ x2r yi2r − |xi yi |r (5.66) i i=1

i=1

 i=1 n n X X ≤ r2 (M1 M2 )2(r−1)  x2i yi2 − i=1

i=1

n X i=1

!2  |xi yi |

.

5.11. AN INEQUALITY VIA JENSEN’S DISCRETE INEQUALITY 173 2. If 0 < m1 ≤ |xi | , 0 < m2 ≤ |yi | , i ∈ {1, . . . , n} and r ∈ (0, 1) , then !2 n n n X X X r yi2r − |xi yi | (5.67) 0≤ x2r i i=1

i=1

≤

r

2

(m1 m2 )2(r−1)

i=1  n n X X 2  xi yi2 − i=1

i=1

n X

!2  |xi yi |

.

i=1

3. If 0 ≤ |xi | ≤ M1 , 0 ≤ |yi | ≤ M2 , i ∈ {1, . . . , n} , then for any natural number k one has n n X X 2k+1 2k+1 0≤ xi |xi | yi2k+1 |yi |2k+1 (5.68) i=1 i=1 n n X X 2k+1 2k+1 2k+1 2k+1 xi |yi | yi |xi | − i=1 i=1 !2 n n n X X X 2(2k+1) 2(2k+1) ≤ xi yi − x2k+1 yi2k+1 i i=1

i=1

i=1  n n X X 2 4k  2 ≤ (2k + 1) (M1 M2 ) xi yi2 − i=1

i=1

n X

!2  |xi yi |

.

i=1

4. If 0 < m1 ≤ xi , 0 < m2 ≤ yi , for any i ∈ {1, . . . , n} , then one has the inequality 2 "X  # 2 n n   X xi xi − ln (5.69) 0≤n ln yi yi i=1 i=1  !2  n n n X X X 1  ≤ x2i yi2 − xi yi  . 2 (m1 m2 ) i=1 i=1 i=1

5.11

An Inequality via Jensen’s Discrete Inequality

The following result holds [9].

174

CHAPTER 5. RELATED INEQUALITIES

¯ = (b1 , . . . , bn ) be two sequences of real Theorem 215 Let ¯ a = (a1 , . . . , an ), b numbers with ai 6= 0, i ∈ {1, . . . , n} . If f : I ⊆ R → R is a convex (concave) function on I and abii ∈ I for each i ∈ {1, . . . , n} , and w ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers, then   Pn bi  Pn  2 w a f i i i=1 ai wi ai bi i=1 Pn f Pn ≤ (≥) . (5.70) 2 2 i=1 wi ai i=1 wi ai Proof. We shall use Jensen’s discrete inequality for convex (concave) functions ! n n 1 X 1 X f pi xi ≤ (≥) pi f (xi ) , (5.71) Pn i=1 Pn i=1 P where pi ≥ 0 with Pn := ni=1 pi > 0 and xi ∈ I for each i ∈ {1, . . . , n} . If in (5.71) we choose xi = abii and pi = wi a2i , then by (5.71) we deduce the desired result (5.70). The following corollary holds [9]. ¯ be sequences of positive real numbers and asCorollary 216 Let ¯ a and b sume that w ¯ is as above. If p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , then one has the inequality n X i=1

!p wi ai bi

≤ (≥)

n X i=1

!p−1 wi a2i

n X

wi a2−p bpi . i

(5.72)

i=1

Proof. Follows by Theorem 215 applied for convex (concave) function f : [0, ∞) → R, f (x) = xp , p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) . Remark 217 If p = 2, then by (5.72) we deduce the (CBS) −inequality.

5.12

An Inequality via Lah-Ribari´ c Inequality

The following counterpart of Jensen’s discrete inequality was obtained in 1973 by Lah and Ribari´c [10].

´ INEQUALITY 5.12. AN INEQUALITY VIA LAH-RIBARIC

175

Lemma 218 Let f : I → R be a convex function, xi ∈ [m, M ] ⊆ I for each i ∈ {1, . . . , n} and p ¯ = (p1 , . . . , pn ) be a positive n−tuple. Then n 1 X pi f (xi ) Pn i=1

≤

1 Pn

M−

Pn

i=1

pi xi

M −m

f (m) +

1 Pn

Pn

i=1

pi xi − m

M −m

f (M ) . (5.73)

Proof. We observe for each i ∈ {1, . . . , n} , that xi =

(M − xi ) m + (xi − m) M . M −m

(5.74)

If in the definition of convexity, i.e., α, β ≥ 0, α + β > 0   αa + βb αf (a) + βf (b) f ≤ α+β α+β

(5.75)

we choose α = M − xi , β = xi − m, a = m and b = M, we deduce, by (5.75), that   (M − xi ) m + (xi − m) M f (xi ) = f (5.76) M −m (M − xi ) f (m) + (xi − m) f (M ) ≤ M −m for each i ∈ {1, . . . , n} . If we multiply (5.76) by pi > 0 and sum over i from 1 to n, we deduce (5.73). The following result holds. ¯ = (b1 , . . . , bn ) be two sequences Theorem 219 Let ¯ a = (a1 , . . . , an ) , b of real numbers with ai 6= 0, i ∈ {1, . . . , n} . If I ⊆ R → R is a convex (concave) function on I and abii ∈ [m, M ] ⊆ I for each i ∈ {1, . . . , n} and w ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers, then   Pn bi 2 i=1 wi ai f ai Pn 2 i=1 wi ai M− ≤ (≥)

Pn wi ai bi Pi=1 n 2 i=1 wi ai

M −m

f (m) +

Pn wi ai bi Pi=1 n 2 i=1 wi ai

−m

M −m

f (M ) . (5.77)

176

CHAPTER 5. RELATED INEQUALITIES

Proof. Follows by Lemma 218 for the choices pi = wi a2i , xi = {1, . . . , n} . The following corollary holds.

bi , ai

i ∈

¯ = (b1 , . . . , bn ) be two sequences of Corollary 220 Let ¯ a = (a1 , . . . , an ) , b positive real numbers and such that 0<m≤

bi ≤ M < ∞ for each i ∈ {1, . . . , n} . ai

(5.78)

If w ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers and p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , then one has the inequality n X

n

wi a2−p bpi + i

i=1

M m (M p−1 − mp−1 ) X wi a2i M −m i=1 n M p − mp X ≤ (≥) wi ai bi . (5.79) M − m i=1

Proof. If we write the inequality (5.77) for the convex (concave) function f (x) = xp , p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , we get n wi ai bi Pn 2−p p Pi=1 M − n 2 b w a i i i i=1 wi ai i=1 Pn ≤ (≥) mp + 2 w a M − m i=1 i i

P

Pn wi ai bi Pi=1 n 2 i=1 wi ai

−m

M −m

Mp

which, after elementary calculations, is equivalent to (5.79). Remark 221 For p = 2, we get n X i=1

wi b2i

+ Mm

n X i=1

wi a2i

≤ (M + m)

n X

wi ai bi

(5.80)

i=1

which is the well known Diaz-Metcalf inequality [11].

5.13

An Inequality via Dragomir-Ionescu Inequality

The following counterpart of Jensen’s inequality was proved in 1994 by S.S. Dragomir and N.M. Ionescu [12].

5.13. AN INEQUALITY VIA DRAGOMIR-IONESCU INEQUALITY 177 ˚ Lemma 222 Let f : I ⊆ R → R be a differentiable convex function Pn on I, ˚ xi ∈I (i ∈ {1, . . . , n}) and pi ≥ 0 (i ∈ {1, . . . , n}) such that Pn := i=1 pi > 0. Then one has the inequality ! n n 1 X 1 X 0≤ pi f (xi ) − f pi xi (5.81) Pn i=1 Pn i=1 n n n 1 X 1 X 1 X 0 ≤ pi xi f (xi ) − pi xi · pi f 0 (xi ) . Pn i=1 Pn i=1 Pn i=1

Proof. Since f is differentiable convex on ˚ I, one has f (x) − f (y) ≥ (x − y) f 0 (y) , for any x, y ∈˚ I. If we choose x = f

1 Pn

(5.82)

Pn

pi xi and y = yk , k ∈ {1, . . . , n} , we get ! ! n n X X 1 1 pi xi − f (yk ) ≥ pi xi − yk f 0 (yk ) . (5.83) Pn i=1 Pn i=1 i=1

Multiplying (5.83) by pk ≥ 0 and summing over k from 1 to n, we deduce the desired result (5.81). The following result holds [9]. ¯ = (b1 , . . . , bn ) be two sequences of Theorem 223 Let ¯ a = (a1 , . . . , an ), b real numbers with ai 6= 0, i ∈ {1, . . . , n} . If f : I ⊆ R → R is a differentiable convex (concave) function on ˚ I and abii ∈˚ I for each i ∈ {1, . . . , n} , and w ¯ = (w1 , . . . , wn ) is a sequence of nonnegative real numbers, then   n n X X bi 2 2 0 ≤ wi ai wi ai f (5.84) ai i=1 i=1 !2  Pn  n X wi ai bi 2 i=1 − wi ai · f Pn 2 i=1 wi ai i=1   n n X X bi 2 0 ≤ wi ai wi ai bi f ai i=1 i=1   n n X X bi 2 0 − wi ai bi wi ai f . a i i=1 i=1

178

CHAPTER 5. RELATED INEQUALITIES

Proof. Follows from Lemma 222 on choosing pi = wi a2i , xi = {1, . . . , n} . The following corollary holds [9].

bi , ai

i ∈

¯ = (b1 , . . . , bn ) be two sequences of Corollary 224 Let ¯ a = (a1 , . . . , an ), b positive real numbers with ai 6= 0, i ∈ {1, . . . , n} . If p ∈ [1, ∞) then one has the inequality 0≤

n X

wi a2i

i=1

" ≤p

n X

bpi − wi a2−p i

n X

i=1

n X

wi a2i

n X

wi a2−p bi − i

i=1

i=1

!2−p

n X

wi a2i

i=1 n X

!p wi ai bi

(5.85)

i=1

wi ai bi

n X

i=1

# wi a3−p bp−1 . i i

i=1

If p ∈ (0, 1) , then n X

0≤ ≤p

!2−p wi a2i

"i=1n X

! wi ai bi

−

i=1

wi ai bi

i=1

5.14

n X

n X

wi a3−p bpi − i

i=1

n X

wi a2i

i=1 n X i=1

wi a2i

n X

wi a2−p bpi i

(5.86)

i=1 n X

# bi . wi a2−p i

i=1

An Inequality via a Refinement of Jensen’s Inequality

We will use the following lemma which contains a refinement of Jensen’s inequality obtained in [13]. Lemma 225 Let f : I ⊆ P R →R be a convex function on the interval I and xi ∈ I, pi ≥ 0 with Pn := ni=1 pi > 0. Then the following inequality holds: f

n 1 X pi xi Pn i=1

!

n X

xi1 + · · · + xik+1 ≤ k+1 pi1 · · · pik f Pn i ,...,i =1 k+1 1 k+1   n 1 X xi1 + · · · + xik ≤ k pi · · · pik f Pn i ,...,i =1 1 k 1

1

k



 (5.87)

5.14. AN INEQUALITY VIA A REFINEMENT OF JENSEN’S INEQUALITY179 ≤ ··· ≤

n 1 X pi f (xi ) , Pn i=1

where k ≥ 1, k ∈ N. Proof. We shall follow the proof in [13]. The first inequality follows by Jensen’s inequality for multiple sums   x +···+x  Pn i1 ik+1 p · · · p ik+1 i1 ,...,ik+1 =1 i1 k+1  Pn f i1 ,...,ik+1 =1 pi1 · · · pik+1  x +···+x  Pn i1 ik+1 i1 ,...,ik+1 =1 pi1 · · · pik+1 f k+1 Pn = i1 ,...,ik+1 =1 pi1 · · · pik+1 since

 x +···+x  i1 ik+1 n p · · · p X ik+1 i1 ,...,ik+1 =1 i1 k+1 Pn = Pnk pi xi i1 ,...,ik+1 =1 pi1 · · · pik+1 i=1

Pn

and

n X

pi1 · · · pik+1 = Pnk+1 .

i1 ,...,ik+1 =1

Now, applying Jensen’s inequality for y1 =

xi1 + xi2 · · · + xik−1 + xik xi + xi3 · · · + xik + xik+1 , y2 = 2 , k k xi + xi1 + xi2 + · · · + xik−1 · · · , yk+1 = k+1 k

we have   y1 + y2 + · · · + yk + yk+1 f (y1 ) + f (y2 ) + · · · + f (yk ) + f (yk+1 ) f ≤ , k+1 k+1 which is equivalent to   xi1 + · · · + xik+1 f k+1  x +x +x +···+x   x +x ···+x +x  i1 i2 ik−1 ik ik+1 i1 i2 ik−1 f + ··· + f k k ≤ . (5.88) k+1

180

CHAPTER 5. RELATED INEQUALITIES

Multiplying (5.88) with the nonnegative real numbers pi1 , . . . , pik+1 and summing over i1 , . . . , ik+1 from 1 to n we deduce n X

 pi1 · · · pik+1 f

i1 ,...,ik+1 =1

xi1 + · · · + xik+1 k+1



n 1  X ≤ k + 1 i ,...,i 1

 pi1 · · · pik+1 f

k+1 =1



xi1 + · · · + xik k

(5.89) 

  xik+1 + xi1 + xi2 + · · · + xik−1  + ··· + pi1 · · · pik+1 f k i1 ,...,ik+1 =1   n X xi1 + · · · + xik = Pn pi1 · · · pik f k i ,...,i =1 n X

1



k

which proves the second part of (5.87). The following result holds. Theorem 226 Let f : I ⊆ R →R be a convex function on the interval I, ¯ a= ¯ (a1 , . . . , an ), b = (b1 , . . . , bn ) real numbers such that ai 6= 0, i ∈ {1, . . . , n} and abii ∈ I, i ∈ {1, . . . , n} . If w ¯ = (w1 , . . . , wn ) are positive real numbers, then  Pn  wi ai bi i=1 f Pn (5.90) 2 i=1 wi ai n X 1 ≤ Pn wi1 · · · wik+1 a2i1 · · · a2ik+1 2 k+1 ( i=1 wi ai ) i1 ,...,ik+1 =1  bi bik+1  1 + · · · + ai aik+1  ×f 1 k+1 1

n X

≤ Pn wi1 · · · wik a2i1 · · · a2ik 2 k ( i=1 wi ai ) i1 ,...,ik =1 b  bik i1 + · · · + ai aik  ×f 1 k

5.14. AN INEQUALITY VIA A REFINEMENT OF JENSEN’S INEQUALITY181 ≤ · · · ≤ Pn

n X

1

i=1

wi a2i

wi a2i f

i=1

  bi . ai

The proof is obvious by Lemma 225 applied for pi = wi a2i , xi = i ∈ {1, . . . , n} . The following corollary holds.

bi , ai

¯ and w Corollary 227 Let ¯ a, b ¯ be sequences of positive real numbers. If p ∈ (−∞, 0) ∪ [1, ∞) (p ∈ (0, 1)) , then one has the inequalities !p n X wi ai bi i=1

Pn   n p−k−1 X bik+1 p bi1 ≤ ( i=1 wi a2i ) 2 2 wi1 · · · wik+1 ai1 · · · aik+1 + ··· + (≥) (k + 1)p a aik+1 i 1 i1 ,...,ik+1 =1 Pn  p n p−k X bi1 bik ≤ ( i=1 wi a2i ) 2 2 wi1 · · · wik ai1 · · · aik + ··· + (≥) kp a aik i 1 i1 ,...,ik =1 !p−1 n n X X ≤ ≤ ··· wi a2i wi a2−p bpi . i (≥) (≥) i=1

i=1

Remark 228 If p = 2, then we deduce the following refinement of the (CBS) − inequality n X i=1

!2 wi ai bi

P 1−k ( ni=1 wi a2i ) ≤ (k + 1)2 P 2−k ( ni=1 wi a2i ) ≤ k2 ≤ ··· ≤ ≤

n X i=1

 2 k+1 k+1 X Y  wi1 · · · wik+1  bi` aij 

n X i1 ,...,ik+1 =1 n X

 2 k k X Y  wi1 · · · wik  bi` aij 

i1 ,...,ik =1

`=1

n 1X wi wj (bi aj + ai bj )2 4 i,j=1

wi a2i

n X i=1

j=1 j6=`

`=1

wi b2i .

j=1 j6=`

182

CHAPTER 5. RELATED INEQUALITIES

5.15

Another Refinement via Jensen’s Inequality

The following refinement of Jensen’s inequality holds (see [15]). Lemma 229 Let f : [a, b] → R beP a differentiable convex function on (a, b) and xi ∈ (a, b) , pi ≥ 0 with Pn := ni=1 pi > 0. Then one has the inequality ! n 1 X pi xi Pn i=1 ! n n 1 X X 1 ≥ pi f (xi ) − f pj xj Pn P n j=1 i=1 ! n n n X X X 1 1 0 1 pj xj · pi xi − pj xj ≥ 0. (5.91) − f Pn Pn j=1 Pn j=1 i=1

n 1 X pi f (xi ) − f Pn i=1

Proof. Since f is differentiable convex on (a, b) , then for each x, y ∈ (a, b), one has the inequality f (x) − f (y) ≥ (x − y) f 0 (y) .

(5.92)

Using the properties of the modulus, we have f (x) − f (y) − (x − y) f 0 (y) = |f (x) − f (y) − (x − y) f 0 (y)| ≥ ||f (x) − f (y)| − |x − y| |f 0 (y)||

(5.93)

for each x, y ∈ (a, b) . P If we choose y = P1n nj=1 pj xj and x = xi , i ∈ {1, . . . , n} , then we have f (xi ) − f

n 1 X pj xj Pn j=1

! −

n 1 X xi − pj xj Pn j=1

! f0

n 1 X pj xj Pn j=1

!

! n X 1 ≥ f (xi ) − f pj xj Pn j=1 ! n n 1 X 0 1 X − xi − pj xj f pj xj (5.94) Pn j=1 Pn j=1

5.15. ANOTHER REFINEMENT VIA JENSEN’S INEQUALITY

183

for any i ∈ {1, . . . , n} . If we multiply (5.94) by pi ≥ 0, sum over i from 1 to n, and divide by Pn > 0, we deduce ! n n 1 X 1 X pi f (xi ) − f pj xj Pn i=1 Pn j=1 ! ! n n n X 1 X 1 1 X − pi xi − pj xj f 0 pj xj Pn i=1 Pn j=1 Pn j=1 ! n n X X 1 1 ≥ pi f (xi ) − f pj xj Pn i=1 Pn j=1 ! n n 1 X 0 1 X pj x j f pj xj − xi − Pn j=1 Pn j=1 ! n n 1 X 1 X pi f (xi ) − f pj xj ≥ Pn P n j=1 i=1 ! n n n X X X 1 1 0 1 pi xi − pj x j · f pj xj . − Pn Pn Pn i=1

Since

j=1

j=1

n n 1 X 1 X pi xi − pj xj Pn i=1 Pn j=1

! = 0,

the inequality (5.91) is proved. In particular, we have the following result for unweighted means. Corollary 230 With the above assumptions for f and xi , one has the inequality   f (x1 ) + · · · + f (xn ) x1 + · · · + xn −f n n n   1 X x1 + · · · + xn ≥ xi − f n n i=1   n n X 0 x1 + · · · + xn 1 X 1 − f · xj ≥ 0. (5.95) xi − n n n i=1

j=1

184

CHAPTER 5. RELATED INEQUALITIES

The following refinement of the (CBS) −inequality holds. Theorem 231 If ai , bi ∈ R, i ∈ {1, . . . , n} , then one has the inequality; n X i=1

a2i

n X

b2i −

i=1

n X

!2 ai bi

i=1

2 2 n X b a i i P 2  P 2 ≥ Pn 2 n n 2 i=1 bi i=1 j=1 aj bj j=1 bj n n n X X X a b −2 ak b k · |bi | bj i i ≥ 0. (5.96) aj b j 1

i=1

k=1

j=1

Proof. If we apply Lemma 229 for f (x) = x2 , we get n 1 X 2 pi xi − Pn i=1

!2 n 1 X pi xi Pn i=1 !2 n n X X 1 1 pi x2i − pj xj ≥ Pn j=1 Pn i=1 n n n 1 X 1 X 1 X pk xk · pi xi − pj xj ≥ 0. (5.97) − 2 Pn Pn Pn j=1 i=1 k=1

If in (5.97), we choose pi = b2i , xi =

ai , bi

i ∈ {1, . . . , n} , we get

Pn 2 P 2 ai ( ni=1 ai bi ) i=1 Pn 2 − Pn 2 ( i=1 b2i ) i=1 bi !2 Pn 2 n X 1 a b a j j Pj=1 ≥ Pn 2 b2i · 2i − n 2 b j i=1 bi i=1 bi j=1 . Pn b2 ai − Pn a b Pn b2 Pn k=1 ak bk i bi j=1 j j j=1 j · i=1 , (5.98) P −2 P n n 2 2 b b i=1 i i=1 i which is clearly equivalent to (5.96).

5.16. AN INEQUALITY VIA SLATER’S RESULT

5.16

185

An Inequality via Slater’s Result ◦

Suppose that I is an interval of real numbers with interior I and f : I → R ◦ is a convex function on I. Then f is continuous on I and has finite left ◦ ◦ and right derivatives at each point of I . Moreover, if x, y ∈I and x < y, then D− f (x) ≤ D+ f (x) ≤ D− f (y) ≤ D+ f (y) which shows that both D− f ◦

and D+ f are nondecreasing functions on I . It is also known that a convex function must be differentiable except for at most countably many points. For a convex function f : I → R, the subdifferentialoff denoted by ∂f ◦ is the set of all functions ϕ : I → [−∞, ∞] such that ϕ I ⊂ R and f (x) ≥ f (a) + (x − a) ϕ (a) for any x, a ∈ I.

(5.99)

It is also well known that if f is convex on I, then ∂f is nonempty, D+ f, D− f ∈ ∂f and if ϕ ∈ ∂f, then D− f (x) ≤ ϕ (x) ≤ D+ f (x)

(5.100)

◦

for every x ∈I . In particular, ϕ is a nondecreasing function. ◦ If f is differentiable convex on I , then ∂f = {f 0 } . The following inequality is well known in the literature as Slater’s inequality [16]. Lemma 232 Let f : I → R be P a nondecreasing P (nonincreasing) convex n function, xi ∈ I, pi ≥ 0 with Pn := i=1 pi > 0 and ni=1 pi ϕ (xi ) 6= 0 where ϕ ∈ ∂f. Then one has the inequality:  Pn  n pi xi ϕ (xi ) 1 X i=1 Pn pi f (xi ) ≤ f . (5.101) Pn i=1 i=1 pi ϕ (xi ) Proof. Firstly, observe that since, for example, f is nondecreasing, then ϕ (x) ≥ 0 for any x ∈ I and thus Pn pi xi ϕ (xi ) Pi=1 ∈ I, (5.102) n i=1 pi ϕ (xi ) since it is a convex combination of xi with the positive weights x ϕ (xi ) Pn i , i = 1, . . . , n. i=1 pi ϕ (xi )

186

CHAPTER 5. RELATED INEQUALITIES

A similar argument applies if f is nonincreasing. Now, if we use the inequality (5.99), we deduce f (x) − f (xi ) ≥ (x − xi ) ϕ (xi ) for any x, xi ∈ I, i = 1, . . . , n.

(5.103)

Multiplying (5.103) by pi ≥ 0 and summing over i from 1 to n, we deduce n n n 1 X 1 X 1 X f (x) − pi f (xi ) ≥ x · pi ϕ (xi ) − pi xi ϕ (xi ) Pn i=1 Pn i=1 Pn i=1

(5.104)

for any x ∈ I. If in (5.104) we choose Pn pi xi ϕ (xi ) , x = Pi=1 n i=1 pi ϕ (xi ) which, we have proved that it belongs to I, we deduce the desired inequality (5.101). If one would like to drop the assumption of monotonicity for the function f, then one can state and prove in a similar way the following result. Lemma P 233 Let f : I → R be a convex function, xi ∈ I, pi ≥ 0 with Pn > 0 and ni=1 pi ϕ (xi ) 6= 0, where ϕ ∈ ∂f. If Pn pi xi ϕ (xi ) Pi=1 ∈ I, (5.105) n i=1 pi ϕ (xi ) then the inequality (5.101) holds. The following result in connection to the (CBS) −inequality holds. Theorem 234 Assume that f : R+ → R is a convex function on R+ := Pn 2  bi  [0, ∞), ai , bi ≥ 0 with ai 6= 0, i ∈ {1, . . . , n} and i=1 ai ϕ ai 6= 0 where ϕ ∈ ∂f. (i) If f is monotonic nondecreasing (nonincreasing) in [0, ∞) then    Pn bi   X n n a b ϕ X i i i=1 ai bi 2 2  ai ϕ ≤ ai · f (5.106) Pn 2  bi   . a i aϕ i=1 i=1 i=1

i

ai

5.17. AN INEQUALITY VIA AN ANDRICA-RAS¸A RESULT

187

(ii) If   bi a b ϕ i=1 i i ai Pn 2  bi  ≥ 0, i=1 ai ϕ ai

Pn

(5.107)

then (5.106) also holds. Remark 235 Consider the function f : [0, ∞) → R, f (x) = xp , p ≥ 1. Then f is convex and monotonic nondecreasing and ϕ (x) = pxp−1 . Applying (5.106), we may deduce the following inequality:

p

n X

!p+1 a3−p bp−1 i i

i=1

≤

n X i=1

a2i

n X

!p a2−p bpi i

(5.108)

i=1

for p ≥ 1, ai , bi ≥ 0, i = 1, . . . , n.

5.17

An Inequality via an Andrica-Ra¸sa Result

The following Jensen type inequality has been obtained in [17] by Andrica and Ra¸sa. Lemma 236 Let f : [a, b] → R be a twice differentiable function and assume that m = inf f 00 (t) > −∞ and M = sup f 00 (t) < ∞. t∈(a,b)

t∈(a,b)

Pn

If xi ∈ [a, b] and pi ≥ 0 (i = 1, . . . , n) with i=1 pi = 1, then one has the inequalities:  ! !2  n n n n X X X X 1  2 m pi xi − pi xi  ≤ pi f (xi ) − f pi xi (5.109) 2 i=1 i=1 i=1 i=1  !2  n n X X 1 ≤ M pi x2i − pi xi  . 2 i=1 i=1

188

CHAPTER 5. RELATED INEQUALITIES

Proof. Consider the auxiliary function fm (t) := f (t) − 21 mt2 . This 00 function is twice differentiable and fm (t) ≥ 0, t ∈ (a, b) , showing that fm is convex. Applying Jensen’s inequality for fm , i.e., n X

pi fm (xi ) ≥ fm

i=1

n X

! pi xi ,

i=1

we deduce, by a simple calculation, the first inequality in (5.109). The second inequality follows in a similar way for the auxiliary function fM (t) = 21 M t2 − f (t) . We omit the details. The above result may be naturally used to obtain the following inequality related to the (CBS) −inequality. ¯ = (b1 , . . . , bn ) be two sequences of Theorem 237 Let ¯ a = (a1 , . . . , an ) , b real numbers with the property that there exists γ, Γ ∈ R such that −∞ ≤ γ ≤

ai ≤ Γ ≤ ∞, for each i ∈ {1, . . . , n} , bi

(5.110)

and bi 6= 0, i = 1, . . . , n. If f : (γ, Γ) → R is twice differentiable and m = inf f 00 (t) and M = sup f 00 (t) , t∈(γ,Γ)

t∈(γ,Γ)

then we have the inequality  n n 1 X 2 X 2 m ai bi − 2 i=1 i=1

n X

!2  ai b i



i=1

!2  P    n n n n X X X ai b i a i 2 2 2 i=1 ≤ bi bi f − bi f Pn 2 b i i=1 bi i=1 i=1  i=1  ! 2 n n n X 1 X 2 X 2 ≤ M ai bi − ai b i  . 2 i=1 i=1 i=1

(5.111)

5.17. AN INEQUALITY VIA AN ANDRICA-RAS¸A RESULT Proof. We may apply Lemma 236 for the choices pi = to get "P  Pn 2 # n a2i ai b i 1 i=1 i=1 m Pn 2 − Pn 2 2 k=1 bk k=1 bk   Pn 2 ai  Pn  i=1 bi f bi ai b i i=1 Pn 2 ≤ − f Pn 2 k=1 bk k=1 bk "P P  2 # n n 2 a a b 1 i i ≤ M Pni=1 i2 − Pi=1 n 2 2 k=1 bk k=1 bk

b2 Pn i

2 k=1 bk

189 and xi =

ai bi

giving the desired result (5.111). The following corollary is a natural consequence of the above theorem. ¯ are sequences of nonnegative real numbers Corollary 238 Assume that ¯ a, b and ai 0<ϕ≤ ≤ Φ < ∞ for each i ∈ {1, . . . , n} . (5.112) bi Then for any p ∈ [1, ∞) one has the inequalities  !2  n n n X X X 1 p (p − 1) ϕp−2  a2i b2i − ai b i  2 i=1 i=1 i=1 ! !p !2−p p n n n n X X X X ≤ b2i api b2−p − b2i ai b i i i=1

i=1

i=1  n n X X 1 ≤ p (p − 1) Φp−2  a2i b2i − 2 i=1 i=1

(5.113)

i=1 n X

!2  ai b i



i=1

if p ∈ [2, ∞) and  !2  n n n X X X 1 p (p − 1) Φp−2  a2i b2i − ai b i  2 i=1 i=1 i=1 ! ! !p p 2−p n n n n X X X X p 2−p ≤ b2i ai b i − b2i ai b i i=1

i=1

i=1

i=1

(5.114)

190

CHAPTER 5. RELATED INEQUALITIES  n n X X 1 p−2  2 ≤ p (p − 1) ϕ ai b2i − 2 i=1 i=1

n X

!2  ai b i



i=1

if p ∈ [1, 2] .

5.18

An Inequality via Jensen’s Result for Double Sums

The following result for convex functions via Jensen’s inequality also holds [18]. Lemma 239 Let f : R → R be a convex (concave) function and x ¯ = (x1 , . . . , xn ) , p ¯ =P (p1 , . . . , pn ) real sequences with the property that pi ≥ 0 (i = 1, . . . , n) and ni=1 pi = 1. Then one has the inequality: # "P Pn 2 n 2 p x − ( p x ) i i i i f Pi=1 Pi=1 2 n n 2p − ( i i i=1 i=1 ipi ) hP i P j−1 p p f (∆x · ∆x ) i j k l 1≤i<j≤n k,l=i ≤ (≥) , (5.115) Pn 2 Pn 2 i=1 i pi − ( i=1 ipi ) where ∆xk := xk+1 − xk (k = 1, . . . , n − 1) is the forward difference. Proof. We have, by Jensen’s inequality for multiple sums that "P # "P # Pn 2 2 n 2 p p (x − x ) p x − ( p x ) i j i j i i i i 1≤i<j≤n f Pi=1 =f P (5.116) Pi=1 2 2 n n 2 p p (j − i) i p − ( ip ) i j i i 1≤i<j≤n i=1 i=1 P  2 (xj −xi )2 1≤i<j≤n pi pj (j − i) (j−i)2  =f P 2 p p (j − i) i j 1≤i<j≤n   P (xj −xi )2 2 1≤i<j≤n pi pj (j − i) f (j−i)2 ≤ =: I. P 2 1≤i<j≤n pi pj (j − i) On the other hand, for j > i, one has xj − xi =

j−1 X k=i

(xk+1 − xk ) =

j−1 X k=i

∆xk

(5.117)

5.18. AN INEQUALITY VIA JENSEN’S RESULT FOR DOUBLE SUMS191 and thus 2

(xj − xi ) =

j−1 X

!2 ∆xk

=

k=i

j−1 X

∆xk · ∆xl .

k,l=i

Applying once more the Jensen inequality for multiple sums, we deduce "

(xj − xi )2 f (j − i)2

" Pj−1

#

k,l=i

=f

∆xk · ∆xl

Pj−1

# ≤ (≥)

(j − i)2

k,l=i

f (∆xk · ∆xl ) (j − i)2

(5.118)

and thus, by (5.118), we deduce Pj−1

f (∆xk ·∆xl )

2 k,l=i 1≤i<j≤n pi pj (j − i) (j−i)2 I ≤ (≥) P 2 1≤i<j≤n pi pj (j − i) hP i P j−1 p p f (∆x · ∆x ) k l 1≤i<j≤n i j k,l=i = , Pn 2 Pn 2 i p − ( ip ) i i i=1 i=1

P

(5.119)

and then, by (5.116) and (5.119), we deduce the desired inequality (5.115). The following inequality connected with the (CBS) −inequality may be stated. Theorem 240 Let f : R → R be a convex (concave) function and ¯ a = ¯ (a1 , . . . , an ) , b = (b1 , . . . , bn ) , w ¯ = (w1 , . . . , wn ) sequences of real numbers such that bi 6= 0, wi ≥ 0 (i = 1, . . . , n) and not all wi are zero. Then one has the inequality "P f

Pn Pn 2 n 2 2 i=1 wi ai i=1 wi bi − ( i=1 wi ai bi ) Pn Pn 2 2 Pn 2 2 i=1 wi bi i=1 i wi bi − ( i=1 iwi bi ) ≤ (≥)

    i ak f ∆ bk · ∆ abll . (5.120) Pn Pn 2 2 Pn 2 2 i=1 wi bi i=1 i wi bi − ( i=1 iwi bi )

2 2 1≤i<j≤n wi wj bi bj

P

#

hP

j−1 k,l=i

Proof. Follows by Lemma 239 on choosing pi = wi b2i and xi = i = 1, . . . , n. We omit the details.

ai , bi

192

5.19

CHAPTER 5. RELATED INEQUALITIES

ˇ Some Inequalities for the Cebyˇ sev Functional

¯ For two sequences of real numbers ¯ a = (a1 , . . . , aP n ) , b = (b1 , . . . , bn ) and n p ¯ = (p1 , . . . , pn ) with pi ≥ 0 (i ∈ {1, . . . , n}) and i=1 pi = 1, consider the ˇ Cebyˇ sev functional n n n X X X
¯ := T p ¯, ¯ a, b p i ai b i − p i ai p i bi . i=1

i=1

(5.121)

i=1

By Korkine’s identity [1, p. 242] one has the representation n
1X ¯ T p ¯, ¯ a, b = pi pj (ai − aj ) (bi − bj ) 2 i,j=1 X = pi pj (aj − ai ) (bj − bi ) .

(5.122)

1≤i<j≤n

Using the (CBS) −inequality for double sums one may state the following result 

¯ 2 ≤ T (¯ ¯ b ¯ , T p ¯, ¯ a, b p, ¯ a, ¯ a) T p ¯ , b, (5.123) where, obviously n 1X pi pj (ai − aj )2 T (¯ p, ¯ a, ¯ a) = 2 i,j=1 X = pi pj (aj − ai )2 .

(5.124)

1≤i<j≤n

The following result holds [14]. ¯ = (b1 , . . . , bn ) are real numLemma 241 Assume that ¯ a = (a1 , . . . , an ) , b bers such that for each i, j ∈ {1, . . . , n} , i < j, one has m (bj − bi ) ≤ aj − ai ≤ M (bj − bi ) ,

(5.125)

where m, M are given real numbers. P If p ¯ = (p1 , . . . , pn ) is a nonnegative sequence with ni=1 pi = 1, then one has the inequality

¯ ≥ T (¯ ¯ b ¯ . (m + M ) T p ¯, ¯ a, b p, ¯ a, ¯ a) + mM T p ¯ , b, (5.126)

ˇ ˇ 5.19. SOME INEQUALITIES FOR THE CEBY SEV FUNCTIONAL 193 Proof. If we use the condition (5.125), we get [M (bi − bj ) − (ai − aj )] [(ai − aj ) − m (bi − bj )] ≥ 0

(5.127)

for i, j ∈ {1, . . . , n} , i < j. If we multiply in (5.127), then, obviously, for any i, j ∈ {1, . . . , n} , i < j we have (aj − ai )2 + mM (bj − bi )2 ≤ (m + M ) (aj − ai ) (bj − bi ) .

(5.128)

Multiplying (5.128) by pi pj ≥ 0, i, j ∈ {1, . . . , n} , i < j, summing over i and j, i < j from 1 to n and using the identities (5.122) and (5.124), we deduce the required inequality (5.125). The following result holds [14]. ¯ p Theorem 242 If ¯ a, b, ¯ are as in Lemma 241 and M ≥ m > 0, then one has the inequality providing a counterpart for (5.123) 
 ¯ 2≥ T p ¯, ¯ a, b


4mM ¯ b ¯ . p, ¯ a, ¯ a) T p ¯ , b, 2 T (¯ (m + M )

(5.129)

Proof. We use the following elementary inequality αx2 +

1 2 y ≥ 2xy, x, y ≥ 0, α > 0 α

(5.130)

to get, for the choices √ 
 1 1 ¯ b ¯ 2 ≥ 0, y = [T (¯ α = mM > 0, x = T p ¯ , b, p, ¯ a, ¯ a)] 2 ≥ 0 the following inequality: √


¯ b ¯ + √ 1 T (¯ p, ¯ a, ¯ a) mM T p ¯ , b, mM 
 1 1 ¯ b ¯ 2 [T (¯ p, ¯ a, ¯ a)] 2 . (5.131) ≥2 T p ¯ , b,

Using (5.130) and (5.131), we deduce

 1 1 (m + M ) ¯ ≥ T p ¯ b ¯ 2 [T (¯ √ T p ¯, ¯ a, b ¯ , b, p, ¯ a, ¯ a)] 2 2 mM which is clearly equivalent to (5.129). The following corollary also holds [14].

194

CHAPTER 5. RELATED INEQUALITIES

Corollary 243 With the assumptions of Theorem 242, we have: 
 1
1 ¯ ¯ b ¯ 2 [T (¯ 0≤ T p ¯ , b, p, ¯ a, ¯ a)] 2 − T p ¯, ¯ a, b √ √ 2 M− m
¯ √ T p ¯, ¯ a, b ≤ 2 mM

(5.132)

and

 ¯ b ¯ − T p ¯ 2 0 ≤ T (¯ p, ¯ a, ¯ a) T p ¯ , b, ¯, ¯ a, b
 (M − m)2  ¯ 2. ≤ T p ¯, ¯ a, b 4mM

(5.133)

The following result is useful in practical applications [14]. Theorem 244 Let f, g : [α, β] → R be continuous on [α, β] and differentiable on (α, β) with g 0 (x) 6= 0 for x ∈ (α, β) . Assume f 0 (x) , x∈(α,β) g 0 (x)

−∞ < γ = inf

f 0 (x) = Γ < ∞. 0 x∈(α,β) g (x) sup

(5.134)

If x ¯ is a real sequence with xi ∈ [α, β] and xi 6= xj for i 6= j and if we denote by f (¯ x) := (f (x1 ) , . . . , f (xn )) , then we have the inequality: (γ + Γ) T (¯ p, f (¯ x) , g (¯ x)) ≥ T (¯ p, f (¯ x) , f (¯ x)) + γΓT (¯ p, g (¯ x) , g (¯ x)) (5.135) P for any p¯ with pi ≥ 0 (i ∈ {1, . . . , n}) , ni=1 pi = 1. Proof. Applying the Cauchy Mean-Value Theorem, there exists ξ ij ∈ (α, β) (i < j) such that
f 0 ξ ij f (xj ) − f (xi )
∈ [γ, Γ] = 0 (5.136) g (xj ) − g (xi ) g ξ ij for i, j ∈ {1, . . . , n} , i < j. Then    f (xj ) − f (xi ) f (xj ) − f (xi ) Γ− − γ ≥ 0, 1 ≤ i < j ≤ n, g (xj ) − g (xi ) g (xj ) − g (xi )

(5.137)

which, by a similar argument to the one in Lemma 241 will give the desired result (5.135). The following corollary is natural [14].

ˇ ˇ 5.20. OTHER INEQUALITIES FOR THE CEBY SEV FUNCTIONAL 195 Corollary 245 With the assumptions in Theorem 244 and if Γ ≥ γ > 0, then one has the inequalities: [T (¯ p, f (¯ x) , g (¯ x))]2 ≥

4γΓ T (¯ p, f (¯ x) , f (¯ x)) T (¯ p, g (¯ x) , g (¯ x)) , (5.138) (γ + Γ)2 1

1

0 ≤ [T (¯ p, f (¯ x) , f (¯ x))] 2 [T (¯ p, g (¯ x) , g (¯ x))] 2 − T (¯ p, f (¯ x) , g (¯ x)) (5.139) √  2 √ Γ− γ √ ≤ T (¯ p, f (¯ x) , g (¯ x)) 2 γΓ and 0 ≤ T (¯ p, f (¯ x) , f (¯ x)) T (¯ p, g (¯ x) , g (¯ x)) − T 2 (¯ p, f (¯ x) , g (¯ x)) ≤

5.20

(5.140)

(Γ − γ)2 2 T (¯ p, f (¯ x) , g (¯ x)) . 4γΓ

ˇ Other Inequalities for the Cebyˇ sev Functional

¯ For two sequences of real numbers ¯ a = (a1 , . . . , aP n ) , b = (b1 , . . . , bn ) and n p ¯ = (p1 , . . . , pn ) with pi ≥ 0 (i ∈ {1, . . . , n}) and i=1 pi = 1, consider the ˇ Cebyˇ sev functional n n n X X
X ¯ = T p ¯, ¯ a, b p i ai b i − p i ai p i bi . i=1

i=1

(5.141)

i=1

By Sonin’s identity [1, p. 246] one has the representation n
X

¯ = ¯ , T p ¯, ¯ a, b pi (ai − An (¯ p, ¯ a)) bi − An p ¯, b i=1

where An (¯ p, ¯ a) :=

n X j=1

p j aj ,

n X
¯ := An p ¯, b p j bj . j=1

(5.142)

196

CHAPTER 5. RELATED INEQUALITIES

Using the (CBS) −inequality for weighted sums, we may state the following result 

¯ 2 ≤ T (¯ ¯ b ¯ , T p ¯, ¯ a, b p, ¯ a, ¯ a) T p ¯ , b, (5.143) where, obviously T (¯ p, ¯ a, ¯ a) =

n X

pi (ai − An (¯ p, ¯ a))2 .

(5.144)

i=1

The following result holds [14]. ¯ = (b1 , . . . , bn ) are real numLemma 246 Assume that ¯ a = (a1 , . . . , an ) , b Pn bers, p ¯ =
(p1 , . . . , pn ) are nonnegative numbers with i=1 pi = 1 and bi 6= ¯ for each i ∈ {1, . . . , n} . If An p ¯, b −∞ < l ≤

ai − An (¯ p, ¯ a)
≤ L < ∞ for all i ∈ {1, . . . , n} , ¯ b i − An p ¯, b

(5.145)

where l, L are given real numbers, then one has the inequality

¯ ≥ T (¯ ¯ b ¯ . (l + L) T p ¯, ¯ a, b p, ¯ a, ¯ a) + LlT p ¯ , b,

(5.146)

Proof. Using (5.145) we have ! ! ai − An (¯ p, ¯ a) ai − An (¯ p, ¯ a)

L− ¯ ¯ −l ≥0 b i − An p ¯, b b i − An p ¯, b

(5.147)

for each i ∈ {1, . . . , n} .

¯ 2 ≥ 0, we get If we multiply (5.147) by bi − An p ¯, b ¯ (ai − An (¯ p, ¯ a))2 + Ll bi − An p ¯, b



2

¯ ≤ (L + l) (ai − An (¯ p, ¯ a)) bi − An p ¯, b





(5.148)

for each i ∈ {1, . . . , n} . Finally, if we multiply (5.148) by pi ≥ 0, sum over i from 1 to n and use the identity (5.142) and (5.144), we obtain (5.146). Using Lemma 246 and a similar argument to that in the previous section, we may state the following theorem [14].

ˇ ˇ 5.21. BOUNDS FOR THE CEBY SEV FUNCTIONAL

197

Theorem 247 With the assumption of Lemma 246 and if L ≥ l > 0, then one has the inequality
  ¯ 2≥ T p ¯, ¯ a, b


4lL ¯ b ¯ . p, ¯ a, ¯ a) T p ¯ , b, 2 T (¯ (L + l)

(5.149)

The following corollary is natural [14]. Corollary 248 With the assumptions in Theorem 247 one has

 1 1  ¯ ¯ b ¯ 2 −T p 0 ≤ [T (¯ p, ¯ a, ¯ a)] 2 T p ¯ , b, ¯, ¯ a, b √ √ 2 L− l
¯ , √ ≤ T p ¯, ¯ a, b 2 lL

(5.150)

and

 ¯ b ¯ − T p ¯ 2 0 ≤ T (¯ p, ¯ a, ¯ a) T p ¯ , b, ¯, ¯ a, b
 (L − l)2  ¯ 2. T p ¯, ¯ a, b ≤ 4lL

5.21

(5.151)

ˇ Bounds for the Cebyˇ sev Functional

The following result holds. ¯ = (b1 , . . . , bn ) (with bi 6= bj for i 6= j) Theorem 249 Let ¯ a = (a1 , . . . , an ) , b be two sequences of real numbers with the property that there exists the real constants m, M such that for any 1 ≤ i < j ≤ n one has m≤

aj − ai ≤ M. bj − bi

(5.152)

Then we have the inequality


¯ b ¯ ≤T p ¯ ≤ MT p ¯ b ¯ , mT p ¯ , b, ¯, ¯ a, b ¯ , b, for any nonnegative sequence p ¯ = (p1 , . . . , pn ) with

Pn

i=1

pi = 1.

(5.153)

198

CHAPTER 5. RELATED INEQUALITIES

Proof. From (5.152), by multiplying with (bj − bi )2 > 0, one has m (bj − bi )2 ≤ (aj − ai ) (bj − bi ) ≤ M (bj − bi )2 for any 1 ≤ i < j ≤ n, giving by multiplication with pi pj ≥ 0, that X X m pi pj (bi − bj )2 ≤ pi pj (aj − ai ) (bj − bi ) 1≤i<j≤n

1≤i<j≤n

≤M

X

pi pj (bi − bj )2 .

1≤i<j≤n

Using Korkine’s identity (see for example Section 5.19), we deduce the desired result (5.153). The following corollary is natural. ¯ in Theorem 249 is strictly inCorollary 250 Assume that the sequence b creasing and there exists m, M such that m≤

∆ak ≤ M, ∆bk

k = 1, . . . , n − 1;

(5.154)

where ∆ak := ak+1 − ak is the forward difference, then (5.153) holds true. Proof. Follows from Theorem 249 on taking into account that for j > i and from (5.154) one has m

j−1 X k=i

∆bk ≤

j−1 X

∆ak ≤ M

k=i

j−1 X

∆bk ,

k=i

giving m (bj − bi ) ≤ aj − ai ≤ M (bj − bi ) . Another possibility is to use functions that generate similar inequalities. Theorem 251 Let f, g : [α, β] → R be continuous on [α, β] and differentiable on (α, β) with g 0 (x) 6= 0 for x ∈ (α, β) . Assume that f 0 (x) −∞ < m = inf , x∈(α,β) g 0 (x)

f 0 (x) sup 0 = M < ∞. x∈(α,β) g (x)

If x ¯ = (x1 , . . . , xn ) is a real sequence with xi ∈ [α, β] and xi 6= xj for i 6= j and if we denote f (¯ x) := (f (x1 ) , . . . , f (xn )) , then we have the inequality mT (¯ p, g (¯ x) , g (¯ x)) ≤ T (¯ p, f (¯ x) , g (¯ x)) ≤ M T (¯ p, g (¯ x) , g (¯ x)) .

(5.155)

ˇ ˇ 5.21. BOUNDS FOR THE CEBY SEV FUNCTIONAL

199

Proof. Applying the Cauchy Mean-Value Theorem, for any j > i there exists ξ ij ∈ (α, β) such that
f 0 ξ ij f (xj ) − f (xi )
∈ [m, M ] . = 0 g (xj ) − g (xi ) g ξ ij Then, by Theorem 249 applied for ai = f (xi ) , bi = g (xi ) , we deduce the desired inequality (5.155). The following inequality related to the (CBS) −inequality holds. Theorem 252 Let ¯ a, x ¯, y ¯ be sequences of real numbers such that xi 6= 0 y and xyii 6= xjj for i 6= j, (i, j ∈ {1, . . . , n}) . If there exist the real numbers γ, Γ such that aj − ai γ ≤ yj ≤ Γ for 1 ≤ i < j ≤ n, (5.156) − xyii xj then we have the inequality  n n X X γ x2i yi2 − ≤

i=1

i=1

n X

n X

x2i

i=1

n X

(5.157)

i=1

ai x i y i −

i=1

n X

ai x2i

i=1

 n n X X 2  ≤Γ xi yi2 − i=1

!2  xi yi 

i=1

n X

n X

xi yi

i=1 !2 

xi yi

.

i=1 x2

Proof. Follows by Theorem 249 on choosing pi = Pn i x2 , bi = k=1 k and M = Γ. We omit the details. The following different approach may be considered as well.

yi , xi

m=γ

¯ = (b1 , . . . , bn ) are sequences Theorem 253 Assume that ¯ a = (a1 , . . . , an ) , b of real ¯ = (p1 , . . . , pn ) is
a sequence of nonnegative real numbers Pnumbers, p P ¯ := n pi bi . If with ni=1 pi = 1 and bi 6= An p ¯, b i=1 −∞ < l ≤

ai − An (¯ p, ¯ a)
≤ L < ∞ for each i ∈ {1, . . . , n} , ¯ b i − An p ¯, b

where l, L are given real numbers, then one has the inequality


¯ b ¯ ≤T p ¯ ≤ LT p ¯ b ¯ . lT p ¯ , b, ¯, ¯ a, b ¯ , b,

(5.158)

(5.159)

200

CHAPTER 5. RELATED INEQUALITIES



¯ 2 > 0, we Proof. From (5.158), by multiplying with bi − An p ¯, b deduce



¯ 2 ≤ (ai − An (¯ ¯ l b i − An p ¯, b p, ¯ a)) bi − An p ¯, b

¯ 2, ≤ L b i − An p ¯, b for any i ∈ {1, . . . , n} . By multiplying with pi ≥ 0, and summing over i from 1 to n, we deduce l

n X i=1

p i b i − An

¯ p ¯, b



2

≤

n X

¯ pi (ai − An (¯ p, ¯ a)) bi − An p ¯, b





i=1

≤L

n X

¯ p i b i − An p ¯, b



2

.

i=1

Using Sonin’s identity (see for example Section 5.20), we deduce the desired result (5.159). The following result in connection with the (CBS) −inequality may be stated. ¯ be sequences of real numbers such that xi 6= 0 and Theorem 254 Let ¯ a,x ¯, b yi y 1 6= Pn x2 An x2 , x for i ∈ {1, . . . , n} . If there exists the real numbers xi i=1 i φ, Φ such that   ai − Pn1 x2 An x2 , a i=1 i  ≤ Φ,  φ≤ (5.160) yi Pn1 2 An x2 , y − x xi i=1 xi   where x2 = (x21 , . . . , x2n ) and yx = xy11 , . . . , xynn , then one has the inequality (5.157). Proof. Follows by Theorem 253 on choosing pi = and L = Φ. We omit the details.

x2 Pn i

i=1

x2i

, bi =

yi , xi

l=φ

Bibliography ´ J.E. PECARI ˇ ´ and A.M. FINK, Classical and New [1] D.S. MITRINOVIC, C Inequalities in Analysis, Kluwer Academic Publishers, 1993. [2] A.M. OSTROWSKI, Vorlesungen u ¨ber Differential und Integralrechnung II, Birkh¨auser, Basel, 1951. [3] S.S. DRAGOMIR, Ostrowski’s inequality in complex inner product spaces, submitted. [4] S.S. DRAGOMIR, Another Ostrowski type inequality, submitted. [5] K. FAN and J. TODD, A determinantal inequality, J. London Math. Soc., 30 (1955), 58-64. [6] S.S. DRAGOMIR and V.M. BOLDEA, Some new remarks on CauchyBuniakowski-Schwartz’s inequality, Coll. Sci. Pap. Faculty of Science Kragujevac, 21 (1999), 33-40. [7] S.S. DRAGOMIR, On Cauchy-Buniakowski-Schwartz’s Inequality for Real Numbers (Romanian), “Caiete Metodico-S¸tiintifice”, No. 57, 1989, pp. 24, Faculty of Mathematics, University of Timi¸soara, Romania. ˇ ´ and S.Z. ˇ ´ A [8] S.S. DRAGOMIR, D.M. MILOSEVI C ARSLANAGIC, Cauchy-Buniakowski-Schwartz inequality for Lipschitzian functions, Coll. Sci. Pap. of the Faculty of Science, Kragujevac, 14 (1991), 2528. [9] S.S. DRAGOMIR, Generalisation of the (CBS) −inequality via Jensen type inequalities, submitted. 201

202

BIBLIOGRAPHY

´ Converse of Jensen’s inequality for convex [10] P. LAH and M. RIBARIC, functions, Univ. Beograd Publ. Elek. Fak. Ser. Math, Fiz., No. 412460, (1973), 201 -205. [11] J.B. DIAZ and F.T. METCALF, Stronger forms of a class of inequalities of G. P´olya-G. Szeg¨o and L.V. Kantorovich, Bull. Amer. Math. Soc., 69 (1963), 415-419. [12] S.S. DRAGOMIR and N.M. IONESCU, Some converse of Jensen’s inequaliy and applications, Anal. Num. Theor. Approx., 23 (1994), 71-78. ˇ ´ and S.S. DRAGOMIR, A refinement of Jensen’s inequal[13] J.E. PECARI C ity and applications, Studia Univ. Babe¸s-Bolyai, Mathematica, 34(1) (1989), 15-19. ˇ [14] S.S. DRAGOMIR, Some inequalities for the Cebyˇ sev functional, The Australian Math. Soc. Gazette., 29 (3) (2002), 164–168. [15] S.S. DRAGOMIR and F.P. SCARMOZZINO, A refinement of Jensen’s discrete inequality for differentiable convex functions, Preprint, RGMIA Res. Rep. Coll., 5(4) (2002), Article 12. [ONLINE: http://rgmia.vu.edu.au/v5n4.html] [16] M.S. SLATER, A companion inequality to Jensen’s inequality, J. Approx. Theory, 32 (1981), 160-166. [17] D. ANDRICA and I. RAS¸A, The Jensen inequality: refinements and applications, Anal. Num. Theor. Approx., (Romania), 14 (1985), 155167. [18] S.S. DRAGOMIR, On an inequality for convex functions, submitted.

Index J−convex function, 18, 19 ˇ Cebyˇ sev functional, 140, 192, 195 ˇ Cebyˇ sev’s inequality, 10, 163, 164 1-norm, 124

complex numbers, 2 complex sequences, 157 counterpart, 107, 116, 117, 119, 120, 123, 126, 129, 132, 135, 140, 143, 146 double sums, 167, 169 for Lipschitzian functions, 171 for weights, 44, 71 functionals associated to, 94 generalisation, 4, 11, 12, 15–17, 19, 27 generalisation via Young’s inequality, 13 inequality related to, 90 real numbers, 1, 4, 29, 56, 73 real sequence, 155 refinement, 10, 42, 50, 54, 55, 57, 80, 82, 86, 88, 91, 92, 181, 184 discrete, 58 type, 23 weights, 9 concave function, 174–177 convex mapping, 166 counterpart inequality, 107, 116, 117, 119–121, 123, 124, 126, 127, 129, 132, 138, 140, 143, 146 complex numbers, 113 Jensen, 174, 176 real numbers, 110

additive, 8, 89, 91, 102, 103, 105, 106 Alzer, 61, 62, 121 Andrica-Badea inequality, 130, 133, 134 arithmetic mean, 49, 89, 94, 165 Arithmetic-Geometric inequality, 11 asynchronous, 10, 163 sequence, 163 barycentric method, 102 Binet-Cauchy’s identity, 3 Callebaut’s inequality, 25, 60 Cassels’ inequality, 101, 106, 107 additive version, 102 refinement, 133, 136 unweighted, 102, 103, 105, 136 Cauchy, 1, 2 Cauchy-Buniakowski-Schwartz, 1 Cauchy-Schwartz, 61 Cauchy Mean-Value Theorem, 194 Cauchy-Buniakowski-Schwartz, 1 Cauchy-Schwartz, 61 CBS-inequality, 1, 12, 13, 15–21, 23, 32, 48, 61, 167, 174 203

204

INDEX ˇ via Cebyˇ sev Functional, 140 via Andrica-Badea, 130 via Diaz-Metcalf, 137

Daykin-Eliezer-Carlitz, 58, 59 De Bruijn’s inequality, 55 Diaz-Metcalf inequality, 137, 139, 176 discriminant, 2, 29, 32 double sums, 167, 169, 192 Dragomir-Ionescu inequality, 176 Dunkl-Williams’ inequality, 60 Euler’s indicator, 21, 170 Fan and Todd inequalities, 161, 162 forward difference, 121, 124 geometric mean, 49, 89, 94, 165 Gr¨ uss inequality, 133 Greub-Rheinboldt inequality, 106 harmonic mean, 49, 165 identity, 3, 4, 57, 60, 83, 86, 121, 141, 196 Binet-Cauchy, 3 Cauchy-Binet, 168 Korkine’s, 192 Lagrange, 2, 4, 61, 83, 87, 168 Sonin, 195 index set mapping, 75, 76, 80, 86, 87, 93 inner product spaces, 2, 60, 84 Jensen’s inequality, 166, 179 counterpart, 176 discrete, 173, 174 counterpart, 174 refinement, 178, 182

Korkine?s identity, 192 Lagrange’s identity, 2, 4, 61, 83, 87, 168 Lah-Ribari´c inequality, 174 Lipschitzian functions, 171 Lupa¸s inequality, 133 matrices, 167 maximum, 102, 131, 132 McLaughlin’s inequality, 57 Milne’s inequality, 59 minimum, 102 monotonicity, 67, 71, 74, 76, 79, 82, 120 multiple sums, 179 n-tuple, 175 non-proportional, 155, 157, 159, 160 Ostrowski’s inequality, 155, 156, 159 p-norm, 127 P´olya-Szeg¨o, 104 P´olya-Szeg¨o’s inequality, 104–106, 137 parameter, 8 Popoviciu?s inequality, 21 positive real numbers, 8, 11, 41, 49, 50, 101, 104, 106, 110, 135, 164, 166, 174, 176, 178, 180, 181 Power functions, 166 power series, 23 proportional, 1, 2, 5, 11 quadratic polynomial, 1, 28 refinement, 47 Cassels’ inequality, 133, 136

INDEX CBS inequality, 10, 42, 50, 54, 55, 57, 61, 80, 82, 86, 88, 91, 92, 181, 184 due to Alzer and Zheng, 61 due to Daykin-Eliezer-Carlitz, 58 in terms of moduli, 35, 41 Jensen’s inequality, 178, 182 non-constant sequences, 50 P´olya-Szeg¨o inequality, 137 Schwitzer’s result, 133 sequence less than the weights, 44 sequence whose norm is 1, 38 via Dunkl-Williams’ Inequality, 60 rotation, 56 Schweitzer inequality, 133 sequence, 49, 57, 68, 77, 110, 120, 123, 141, 156 asynchronous, 163 complex, 157 complex numbers, 2, 12, 30, 61, 107, 121, 124, 127, 137, 156, 159, 160 decreasing nonnegative, 120 natural numbers, 170 nonnegative, 7, 9, 89, 120, 163, 192 nonnegative numbers, 10 nonnegative real numbers, 13, 25, 26, 44, 71, 73, 101, 106, 121, 124, 127, 130, 135, 174– 177 nonzero complex numbers, 61 nonzero real numbers, 20 positive, 117

205 positive and real numbers, 5 positive numbers, 4, 58 positive real numbers, 8, 11, 41, 50, 101, 104, 106, 110, 135, 164, 166, 174, 176, 178, 181 real, 155, 163, 194 real numbers, 4, 6–9, 20, 23, 24, 27, 35–38, 40–44, 46– 50, 55–57, 71, 73, 83, 119, 124, 126, 129, 130, 132, 133, 140, 143, 147, 155, 159, 161, 162, 167–172, 174, 175, 177, 192, 195 synchronous, 164 set-additive, 93, 95 set-superadditive, 94–96, 98 set-superadditivity, 94 set-supermultiplicative, 94 Shisha-Mond inequalities, 116 Sonin?s identity, 195 strong monotonicity, 79, 82 strong superadditivity, 78, 80 sup-norm, 121 superadditivity, 73, 75, 78, 80, 83, 86, 87 supermultiplicative, 89 supermultiplicity, 89, 93 synchronous, 164 sequence, 164 triangle inequality, 4, 122 Wagner?s inequality, 29 Wagners inequality, 27 weights, 9, 44, 71, 73, 78, 89 Young’s Inequality, 10, 12, 13 Zagier inequalities, 118, 119

206 Zheng, 61, 62

INDEX