Old and New Inequalities

a

1

2

b

Preface This work blends together classic inequality results with brand new problems, some of which devised only a few days ago. What could be special about it when so many inequality problem books have already been written? We strongly believe that even if the topic we plunge into is so general and popular our book is very different. Of course, it is quite easy to say this, so we will give some supporting arguments. This book contains a large variety of problems involving inequalities, most of them difficult, questions that became famous in competitions because of their beauty and difficulty. And, even more importantly, throughout the text we employ our own solutions and propose a large number of new original problems. There are memorable problems in this book and memorable solutions as well. This is why this work will clearly appeal to students who are used to use Cauchy-Schwarz as a verb and want to further improve their algebraic skills and techniques. They will find here tough problems, new results, and even problems that could lead to research. The student who is not as keen in this field will also be exposed to a wide variety of moderate and easy problems, ideas, techniques, and all the ingredients leading to a good preparation for mathematical contests. Some of the problems we chose to present are known, but we have included them here with new solutions which show the diversity of ideas pertaining to inequalities. Anyone will find here a challenge to prove his or her skills. If we have not convinced you, then please take a look at the last problems and hopefully you will agree with us. Finally, but not in the end, we would like to extend our deepest appreciation to the proposers of the problems featured in this book and to apologize for not giving all complete sources, even though we have given our best. Also, we would like to thank Marian Tetiva, Dung Tran Nam, Constantin T˘an˘asescu, C˘alin Popa and Valentin Vornicu for the beautiful problems they have given us and for the precious comments, to Cristian Bab˘ a, George Lascu and C˘alin Popa, for typesetting and for the many pertinent observations they have provided.

The authors

3

Contents Preface

3

Chapter 1. Problems

7

Chapter 2. Solutions

25

Glossary

121

Further reading

127

5

CHAPTER 1

Problems

7

8

Problems

1. Prove that the inequality

√ p p 3 2 b2 + (1 − c)2 + c2 + (1 − a)2 ≥ 2 holds for arbitrary real numbers a, b, c. p

a2 + (1 − b)2 +

K¨ omal 2. [ Dinu S ¸ erb˘ anescu ] If a, b, c ∈ (0, 1) prove that p √ abc + (1 − a)(1 − b)(1 − c) < 1. Junior TST 2002, Romania 3. [ Mircea Lascu ] Let a, b, c be positive real numbers such that abc = 1. Prove that √ √ b+c c+a a+b √ √ + √ + √ ≥ a + b + c + 3. a c b Gazeta Matematic˘ a 4. If the equation x4 + ax3 + 2x2 + bx + 1 = 0 has at least one real root, then a + b2 ≥ 8. Tournament of the Towns, 1993 2

5. Find the maximum value of the expression x3 + y 3 + z 3 − 3xyz where x2 + y 2 + z 2 = 1 and x, y, z are real numbers. 6. Let a, b, c, x, y, z be positive real numbers such that x + y + z = 1. Prove that p ax + by + cz + 2 (xy + yz + zx)(ab + bc + ca) ≤ a + b + c. Ukraine, 2001 7. [ Darij Grinberg ] If a, b, c are three positive real numbers, then a 2

(b + c)

+

b 2

(c + a)

+

c 2

(a + b)

≥

9 . 4 (a + b + c)

8. [ Hojoo Lee ] Let a, b, c ≥ 0. Prove that p p p a4 + a2 b2 + b4 + b4 + b2 c2 + c4 + c4 + c2 a2 + a4 ≥ p p p ≥ a 2a2 + bc + b 2b2 + ca + c 2c2 + ab. Gazeta Matematic˘ a 9. If a, b, c are positive real numbers such that abc = 2, then √ √ √ a3 + b3 + c3 ≥ a b + c + b c + a + c a + b.

Old and New Inequalities

9

When does equality hold? JBMO 2002 Shortlist 10. [ Ioan Tomescu ] Let x, y, z > 0. Prove that xyz 1 ≤ 4. (1 + 3x)(x + 8y)(y + 9z)(z + 6) 7 When do we have equality? Gazeta Matematic˘ a 11. [ Mihai Piticari, Dan Popescu ] Prove that 5(a2 + b2 + c2 ) ≤ 6(a3 + b3 + c3 ) + 1, for all a, b, c > 0 with a + b + c = 1. 12. [ Mircea Lascu ] Let x1 , x2 , . . . , xn ∈ R, n ≥ 2 and a > 0 such that  x1 + a2 2a 2 2 2 x2 + . . . + xn = a and x1 + x2 + . . . + xn ≤ . Prove that xi ∈ 0, , for all n−1 n i ∈ {1, 2, . . . , n}. 13. [ Adrian Zahariuc ] Prove that for any a, b, c ∈ (1, 2) the following inequality holds √ √ √ b a a c c b √ + √ √ √ + √ √ ≥ 1. 4b c − c a 4c a − a b 4a b − b c 14. For positive real numbers a, b, c such that abc ≤ 1, prove that c a b + + ≥ a + b + c. b c a

15. [ Vasile Cˆırtoaje, Mircea Lascu ] Let a, b, c, x, y, z be positive real numbers such that a + x ≥ b + y ≥ c + z and a + b + c = x + y + z. Prove that ay + bx ≥ ac + xz.

16. [ Vasile Cˆırtoaje, Mircea Lascu ] Let a, b, c be positive real numbers so that abc = 1. Prove that 6 3 1+ ≥ . a+b+c ab + ac + bc Junior TST 2003, Romania 17. Let a, b, c be positive real numbers. Prove that a3 b3 c3 a2 b2 c2 + 2+ 2 ≥ + + . 2 b c a b c a JBMO 2002 Shortlist

10

Problems

18. Prove that if n > 3 and x1 , x2 , . . . , xn > 0 have product 1, then 1 1 1 + + ··· + > 1. 1 + x1 + x1 x2 1 + x2 + x2 x3 1 + xn + xn x1 Russia, 2004 19. [ Marian Tetiva ] Let x, y, z be positive real numbers satisfying the condition x2 + y 2 + z 2 + 2xyz = 1. Prove that 1 a) xyz ≤ ; 8 3 b) x + y + z ≤ ; 2 3 c) xy + xz + yz ≤ ≤ x2 + y 2 + z 2 ; 4 1 d) xy + xz + yz ≤ + 2xyz. 2 20. [ Marius Olteanu ] Let x1 , x2 , x3 , x4 , x5 ∈ R so that x1 + x2 + x3 + x4 + x5 = 0. Prove that | cos x1 | + | cos x2 | + | cos x3 | + | cos x4 | + | cos x5 | ≥ 1. Gazeta Matematic˘ a 21. [ Florina Cˆ arlan, Marian Tetiva ] Prove that if x, y, z > 0 satisfy the condition

then xy + xz + yz ≥ 3 +

√

x + y + z = xyz p √ x2 + 1 + y 2 + 1 + z 2 + 1.

22. [ Laurent¸iu Panaitopol ] Prove that 1 + y2 1 + z2 1 + x2 + + ≥ 2, 1 + y + z2 1 + z + x2 1 + x + y2 for any real numbers x, y, z > −1. JBMO, 2003 23. Let a, b, c > 0 with a + b + c = 1. Show that a2 + b b2 + c c2 + a + + ≥ 2. b+c c+a a+b

24. Let a, b, c ≥ 0 such that a4 + b4 + c4 ≤ 2(a2 b2 + b2 c2 + c2 a2 ). Prove that a2 + b2 + c2 ≤ 2(ab + bc + ca). Kvant, 1988

Old and New Inequalities

11

25. Let n ≥ 2 and x1 , ..., xn be positive real numbers satisfying 1 1 1 1 + + ... + = . x1 + 1998 x2 + 1998 xn + 1998 1998 Prove that

√ n

x1 x2 ...xn ≥ 1998. n−1 Vietnam, 1998

26. [ Marian Tetiva ] Consider positive real numbers x, y, z so that x2 + y 2 + z 2 = xyz. Prove the following inequalities a) xyz ≥ 27; b) xy + xz + yz ≥ 27; c) x + y + z ≥ 9; d) xy + xz + yz ≥ 2 (x + y + z) + 9. 27. Let x, y, z be positive real numbers with sum 3. Prove that √ √ √ x + y + z ≥ xy + yz + zx. Russia, 2002 28. [ D. Olteanu ] Let a, b, c be positive real numbers. Prove that a+b a b+c b c+a c 3 · + · + · ≥ . b + c 2a + b + c c + a 2b + c + a a + b 2c + a + b 4 Gazeta Matematic˘ a 29. For any positive real numbers a, b, c show that the following inequality holds c c+a a+b b+c a b + + ≥ + + . b c a c+b a+c b+a India, 2002 30. Let a, b, c be positive real numbers. Prove that a3 b3 c3 3(ab + bc + ca) + + ≥ . 2 2 2 2 2 2 b − bc + c c − ac + a a − ab + b a+b+c Proposed for the Balkan Mathematical Olympiad 31. [ Adrian Zahariuc ] Consider the pairwise distinct integers x1 , x2 , . . . , xn , n ≥ 0. Prove that x21 + x22 + · · · + x2n ≥ x1 x2 + x2 x3 + · · · + xn x1 + 2n − 3.

12

Problems

32. [ Murray Klamkin ] Find the maximum value of the expression x21 x2 + x22 x3 + · · · + x2n−1 xn + x2n x1 when x1 , x2 , . . . , xn−1 , xn ≥ 0 add up to 1 and n > 2. Crux Mathematicorum 33. Find the maximum value of the constant c such that for any x1 , x2 , . . . , xn , · · · > 0 for which xk+1 ≥ x1 + x2 + · · · + xk for any k, the inequality √ √ √ √ x1 + x2 + · · · + xn ≤ c x1 + x2 + · · · + xn also holds for any n. IMO Shortlist, 1986 34. Given are positive real numbers a, b, c and x, y, z, for which a + x = b + y = c + z = 1. Prove that 1 1 1 + + ≥ 3. (abc + xyz) ay bz cx Russia, 2002 35. [ Viorel Vˆ ajˆ aitu, Alexandru Zaharescu ] Let a, b, c be positive real numbers. Prove that ab bc ca 1 + + ≤ (a + b + c). a + b + 2c b + c + 2a c + a + 2b 4 Gazeta Matematic˘ a 36. Find the maximum value of the expression a3 (b + c + d) + b3 (c + d + a) + c3 (d + a + b) + d3 (a + b + c) where a, b, c, d are real numbers whose sum of squares is 1. 37. [ Walther Janous ] Let x, y, z be positive real numbers. Prove that x y z p p p + + ≤ 1. x + (x + y)(x + z) y + (y + z)(y + x) z + (z + x)(z + y) Crux Mathematicorum 38. Suppose that a1 < a2 < ... < an are real numbers for some integer n ≥ 2. Prove that a1 a42 + a2 a43 + ... + an a41 ≥ a2 a41 + a3 a42 + ... + a1 a4n . Iran, 1999 39. [ Mircea Lascu ] Let a, b, c be positive real numbers. Prove that   b+c c+a a+b a b c + + ≥4 + + . a b c b+c c+a a+b

Old and New Inequalities

13

40. Let a1 , a2 , . . . , an > 1 be positive integers. Prove that at least one of the √ √ √ √ √ numbers a1 a2 , a2 a3 , . . . , an−1 an , an a1 is less than or equal to 3 3. Adapted after a well-known problem 41. [ Mircea Lascu, Marian Tetiva ] Let x, y, z be positive real numbers which satisfy the condition xy + xz + yz + 2xyz = 1. Prove that the following inequalities hold 1 a) xyz ≤ ; 8 3 b) x + y + z ≥ ; 2 1 1 1 c) + + ≥ 4 (x + y + z); x y z 2 1 1 1 (2z − 1) d) + + − 4 (x + y + z) ≥ , where z = max {x, y, z}. x y z (z (2z + 1)) 42. [ Manlio Marangelli ] Prove that for any positive real numbers x, y, z, 3(x2 y + y 2 z + z 2 x)(xy 2 + yz 2 + zx2 ) ≥ xyz(x + y + z)3 .

43. [ Gabriel Dospinescu ] Prove that if a, b, c are real numbers such that max{a, b, c} − min{a, b, c} ≤ 1, then 1 + a3 + b3 + c3 + 6abc ≥ 3a2 b + 3b2 c + 3c2 a

44. [ Gabriel Dospinescu ] Prove that for any positive real numbers a, b, c we have       a2 b2 c2 1 1 1 27 + 2 + 2+ 2+ ≥ 6(a + b + c) + + . bc ca ab a b c 45. Let a0 =

1 a2 1 and ak+1 = ak + k . Prove that 1 − < an < 1. 2 n n TST Singapore

46. [ C˘ alin Popa ] Let a, b, c be positive real numbers, with a, b, c ∈ (0, 1) such that ab + bc + ca = 1. Prove that   a b c 3 1 − a2 1 − b2 1 − c2 + + ≥ + + . 1 − a2 1 − b2 1 − c2 4 a b c 47. [ Titu Andreescu, Gabriel Dospinescu ] Let x, y, z ≤ 1 and x + y + z = 1. Prove that 1 1 1 27 + + ≤ . 2 2 2 1+x 1+y 1+z 10

14

Problems

48. [ Gabriel Dospinescu ] Prove that if

√

x+

√

y+

√

z = 1, then

(1 − x)2 (1 − y)2 (1 − z)2 ≥ 215 xyz(x + y)(y + z)(z + x) . 49. Let x, y, z be positive real numbers such that xyz = x + y + z + 2. Prove that (1) xy + yz + zx ≥ 2(x + y + z); √ √ 3√ √ (2) x + y + z ≤ xyz. 2 50. Prove that if x, y, z are real numbers such that x2 + y 2 + z 2 = 2, then x + y + z ≤ xyz + 2. IMO Shortlist, 1987 51. [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x1 , x2 , . . . , xn ∈ (0, 1) and for any permutation σ of the set {1, 2, . . . , n}, we have the inequality   n X ! xi   n n X X   1 1 i=1  ≥ . 1 + n  · 1 − xi 1 − xi · xσ(i)   i=1 i=1

52. Let x1 , x2 , . . . , xn be positive real numbers such that

n X i=1

that

n X √ i=1

xi ≥ (n − 1)

1 = 1. Prove 1 + xi

n X 1 √ . xi i=1

Vojtech Jarnik 53. [ Titu Andreescu ] Let n > 3 and a1 , a2 , . . . , an be real numbers such that a1 + a2 + . . . + an ≥ n and a21 + a22 + . . . + a2n ≥ n2 . Prove that max{a1 , a2 , . . . , an } ≥ 2. USAMO, 1999 54. [ Vasile Cˆırtoaje ] If a, b, c, d are positive real numbers, then b−c c−d d−a a−b + + + ≥ 0. b+c c+d d+a a+b

55. If x and y are positive real numbers, show that xy + y x > 1. France, 1996

Old and New Inequalities

15

56. Prove that if a, b, c > 0 have product 1, then (a + b)(b + c)(c + a) ≥ 4(a + b + c − 1). MOSP, 2001 57. Prove that for any a, b, c > 0, (a2 + b2 + c2 )(a + b − c)(b + c − a)(c + a − b) ≤ abc(ab + bc + ca).

58. [ D.P.Mavlo ] Let a, b, c > 0. Prove that 3+a+b+c+

1 1 1 a b c (a + 1)(b + 1)(c + 1) + + + + + ≥3 . a b c b c a 1 + abc Kvant, 1988

59. [ Gabriel Dospinescu ] Prove that for any positive real numbers x1 , x2 , . . . , xn with product 1 we have the inequality !n n n n Y X X 1 n n n · (xi + 1) ≥ xi + . x i=1 i=1 i=1 i

60. Let a, b, c, d > 0 such that a + b + c = 1. Prove that   d 1 1 3 3 3 , + . a + b + c + abcd ≥ min 4 9 27 Kvant, 1993 61. Prove that for any real numbers a, b, c we have the inequality X

(1 + a2 )2 (1 + b2 )2 (a − c)2 (b − c)2 ≥ (1 + a2 )(1 + b2 )(1 + c2 )(a − b)2 (b − c)2 (c − a)2 . AMM

62. [ Titu Andreescu, Mircea Lascu ] Let α, x, y, z be positive real numbers such that xyz = 1 and α ≥ 1. Prove that xα yα zα 3 + + ≥ . y+z z+x x+y 2

y12

63. Prove that for any real numbers x1 , . . . , xn , y1 , . . . , yn such that x21 +· · ·+x2n = + · · · + yn2 = 1, ! n X 2 (x1 y2 − x2 y1 ) ≤ 2 1 − xk yk . k=1

Korea, 2001

16

Problems

64. [ Laurent¸iu Panaitopol ] Let a1 , a2 , . . . , an be pairwise distinct positive integers. Prove that a21 + a22 + · · · + a2n ≥

2n + 1 (a1 + a2 + · · · + an ). 3 TST Romania

65. [ C˘ alin Popa ] Let a, b, c be positive real numbers such that a + b + c = 1. Prove that √ √ √ √ c a a b 3 3 b c √ √ √ + √ + √ ≥ . √ 4 a( 3c + ab) b( 3a + bc) c( 3b + ca)

66. [ Titu Andreescu, Gabriel Dospinescu ] Let a, b, c, d be real numbers such that (1 + a2 )(1 + b2 )(1 + c2 )(1 + d2 ) = 16. Prove that −3 ≤ ab + bc + cd + da + ac + bd − abcd ≤ 5.

67. Prove that (a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca) for any positive real numbers a, b, c. APMO, 2004 68. [ Vasile Cˆırtoaje ] Prove that if 0 < x ≤ y ≤ z and x + y + z = xyz + 2, then a) (1 − xy)(1 − yz)(1 − xz) ≥ 0; 32 b) x2 y ≤ 1, x3 y 2 ≤ . 27 69. [ Titu Andreescu ] Let a, b, c be positive real numbers such that a+b+c ≥ abc. Prove that at least two of the inequalities 2 3 6 2 3 6 2 3 6 + + ≥ 6, + + ≥ 6, + + ≥ 6, a b c b c a c a b are true. TST 2001, USA 70. [ Gabriel Dospinescu, Marian Tetiva ] Let x, y, z > 0 such that x + y + z = xyz. Prove that

√ (x − 1) (y − 1) (z − 1) ≤ 6 3 − 10.

Old and New Inequalities

17

71. [ Marian Tetiva ] Prove that for any positive real numbers a, b, c, 3 a − b3 b3 − c3 c3 − a3 (a − b)2 + (b − c)2 + (c − a)2 . a+b + b+c + c+a ≤ 4 Moldova TST, 2004 72. [ Titu Andreescu ] Let a, b, c be positive real numbers. Prove that (a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a + b + c)3 . USAMO, 2004 73. [ Gabriel Dospinescu ] Let n > 2 and x1 , x2 , . . . , xn > 0 such that ! n ! n X X 1 xk = n2 + 1. xk k=1

k=1

Prove that n X k=1

! x2k

·

n X 1 x2k

k=1

! > n2 + 4 +

2 . n(n − 1)

74. [ Gabriel Dospinescu, Mircea Lascu, Marian Tetiva ] Prove that for any positive real numbers a, b, c, a2 + b2 + c2 + 2abc + 3 ≥ (1 + a)(1 + b)(1 + c).

75. [ Titu Andreescu, Zuming Feng ] Let a, b, c be positive real numbers. Prove that (2b + a + c)2 (2c + a + b)2 (2a + b + c)2 + + ≤ 8. 2a2 + (b + c)2 2b2 + (a + c)2 2c2 + (a + b)2 USAMO, 2003 76. Prove that for any positive real numbers x, y and any positive integers m, n, (n−1)(m−1)(xm+n +y m+n )+(m+n−1)(xm y n +xn y m ) ≥ mn(xm+n−1 y+y m+n−1 x). Austrian-Polish Competition, 1995 77. Let a, b, c, d, e be positive real numbers such that abcde = 1. Prove that a + abc b + bcd c + cde d + dea e + eab 10 + + + + ≥ . 1 + ab + abcd 1 + bc + bcde 1 + cd + cdea 1 + de + deab 1 + ea + eabc 3 Crux Mathematicorum

18

Problems

 π 78. [ Titu Andreescu ] Prove that for any a, b, c, ∈ 0, the following inequality 2 holds sin a · sin(a − b) · sin(a − c) sin b · sin(b − c) · sin(b − a) sin c · sin(c − a) · sin(c − b) + + ≥ 0. sin(b + c) sin(c + a) sin(a + b) TST 2003, USA 79. Prove that if a, b, c are positive real numbers then, p p p p a4 + b4 + c4 + a2 b2 + b2 c2 + c2 a2 ≥ a3 b + b3 c + c3 a + ab3 + bc3 + ca3 . KMO Summer Program Test, 2001 80. [ Gabriel Dospinescu, Mircea Lascu ] For a given n > 2 find the smallest constant kn with the property: if a1 , . . . , an > 0 have product 1, then a1 a2 a2 a3 an a1 ≤ kn . + + ··· + 2 (a21 + a2 )(a22 + a1 ) (a22 + a3 )(a23 + a2 ) (an + a1 )(a21 + an )

81. [ Vasile Cˆırtoaje ] For any real numbers a, b, c, x, y, z prove that the inequality holds p 2 ax + by + cz + (a2 + b2 + c2 )(x2 + y 2 + z 2 ) ≥ (a + b + c)(x + y + z). 3 Kvant, 1989 82. [ Vasile Cˆırtoaje ] Prove that the sides a, b, c of a triangle satisfy the inequality     b c c a a b + + −1 ≥2 + + . 3 b c a a b c 83. [ Walther Janous ] Let n > 2 and let x1 , x2 , . . . , xn > 0 add up to 1. Prove that  Y  n  n  Y 1 n − xi 1+ ≥ . xi 1 − xi i=1 i=1 Crux Mathematicorum 84. [ Vasile Cˆırtoaje, Gheorghe Eckstein ] Consider positive real numbers x1 , x2 , ..., xn such that x1 x2 ...xn = 1. Prove that 1 1 1 + + ... + ≤ 1. n − 1 + x1 n − 1 + x2 n − 1 + xn TST 1999, Romania 85. [ Titu Andreescu ] Prove that for any nonnegative real numbers a, b, c such that a2 + b2 + c2 + abc = 4 we have 0 ≤ ab + bc + ca − abc ≤ 2. USAMO, 2001

Old and New Inequalities

19

86. [ Titu Andreescu ] Prove that for any positive real numbers a, b, c the following inequality holds √ √ √ √ √ √ a+b+c √ 3 − abc ≤ max{( a − b)2 , ( b − c)2 , ( c − a)2 }. 3 TST 2000, USA 87. [ Kiran Kedlaya ] Let a,b,c be positive real numbers. Prove that r √ √ a+b a+b+c a + ab + 3 abc 3 ≤ a· · . 3 2 3 88. Find the greatest constant k such that for any positive integer n which is not √ √ a square, |(1 + n) sin(π n)| > k. Vietnamese IMO Training Camp, 1995 89. [ Dung Tran Nam ] Let x, y, z > 0 such that (x + y + z)3 = 32xyz. Find the x4 + y 4 + z 4 minimum and maximum of . (x + y + z)4 Vietnam, 2004 90. [ George Tsintifas ] Prove that for any a, b, c, d > 0, (a + b)3 (b + c)3 (c + d)3 (d + a)3 ≥ 16a2 b2 c2 d2 (a + b + c + d)4 . Crux Mathematicorum 91. [ Titu Andreescu, Gabriel Dospinescu ] Find the maximum value of the expression (bc)n (ca)n (ab)n + + 1 − ab 1 − bc 1 − ca where a, b, c are nonnegative real numbers which add up to 1 and n is some positive integer. 92. Let a, b, c be positive real numbers. Prove that 1 1 1 3 √ + + ≥ √ . 3 a(1 + b) b(1 + c) c(1 + a) abc(1 + 3 abc)

93. [ Dung Tran Nam ] Prove that for any real numbers a, b, c such that a2 + b2 + c2 = 9, 2(a + b + c) − abc ≤ 10. Vietnam, 2002

20



Problems

94. [ Vasile Cˆırtoaje ] Let a, b, c be positive real numbers. Prove that         1 1 1 1 1 1 a+ −1 b+ −1 + b+ −1 c+ −1 + c+ −1 a + − 1 ≥ 3. b c c a a b

95. [ Gabriel Dospinescu ] Let n be an integer greater than 2. Find the greatest real number mn and the least real number Mn such that for any positive real numbers x1 , x2 , . . . , xn (with xn = x0 , xn+1 = x1 ), mn ≤

n X i=1

xi ≤ Mn . xi−1 + 2(n − 1)xi + xi+1

96. [ Vasile Cˆırtoaje ] If x, y, z are positive real numbers, then 1 1 9 1 + 2 + 2 ≥ . x2 + xy + y 2 y + yz + z 2 z + zx + z 2 (x + y + z)2 Gazeta Matematic˘ a 97. [ Vasile Cˆırtoaje ] For any a, b, c, d > 0 prove that 2(a3 + 1)(b3 + 1)(c3 + 1)(d3 + 1) ≥ (1 + abcd)(1 + a2 )(1 + b2 )(1 + c2 )(1 + d2 ). Gazeta Matematic˘ a 98. Prove that for any real numbers a, b, c, (a + b)4 + (b + c)4 + (c + a)4 ≥

4 4 (a + b4 + c4 ). 7 Vietnam TST, 1996

99. Prove that if a, b, c are positive real numbers such that abc = 1, then 1 1 1 1 1 1 + + ≤ + + . 1+a+b 1+b+c 1+c+a 2+a 2+b 2+c Bulgaria, 1997 1 2 3 100. [ Dung Tran Nam ] Find the minimum value of the expression + + a b c where a, b, c are positive real numbers such that 21ab + 2bc + 8ca ≤ 12. Vietnam, 2001 101. [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x, y, z, a, b, c > 0 such that xy + yz + zx = 3, a b c (y + z) + (z + x) + (x + y) ≥ 3. b+c c+a a+b

Old and New Inequalities

21

102. Let a, b, c be positive real numbers. Prove that (b + c − a)2 (c + a − b)2 (a + b − c)2 3 + + ≥ . 2 2 2 2 2 2 (b + c) + a (c + a) + b (a + b) + c 5 Japan, 1997 103. [ Vasile Cˆırtoaje, Gabriel Dospinescu ] Prove that if a1 , a2 , . . . , an ≥ 0 then n  a1 + a2 + · · · + an−1 − an an1 + an2 + · · · + ann − na1 a2 . . . an ≥ (n − 1) n−1 where an is the least among the numbers a1 , a2 , . . . , an . 104. [ Turkevici ] Prove that for all positive real numbers x, y, z, t, x4 + y 4 + z 4 + t4 + 2xyzt ≥ x2 y 2 + y 2 z 2 + z 2 t2 + t2 x2 + x2 z 2 + y 2 t2 . Kvant 105. Prove that for any real numbers a1 , a2 , . . . , an the following inequality holds n X i=1

!2 ai

≤

n X

ij ai aj . i + j−1 i,j=1

106. Prove that if a1 , a2 , . . . , an , b1 , . . . , bn are real numbers between 1001 and 2002, inclusively, such that a21 + a22 + · · · + a2n = b21 + b22 + · · · + b2n , then we have the inequality
a31 a3 a3 17 2 a1 + a22 + · · · + a2n . + 2 + ··· + n ≤ b1 b2 bn 10 TST Singapore 107. [ Titu Andreescu, Gabriel Dospinescu ] Prove that if a, b, c are positive real numbers which add up to 1, then (a2 + b2 )(b2 + c2 )(c2 + a2 ) ≥ 8(a2 b2 + b2 c2 + c2 a2 )2 .

108. [ Vasile Cˆırtoaje ] If a, b, c, d are positive real numbers such that abcd = 1, then 1 1 1 1 + + + ≥ 1. (1 + a)2 (1 + b)2 (1 + c)2 (1 + d)2 Gazeta Matematic˘ a 109. [ Vasile Cˆırtoaje ] Let a, b, c be positive real numbers. Prove that b2

a2 b2 c2 a b c + 2 + 2 ≥ + + . 2 2 +c c +a a + b2 b+c c+a a+b Gazeta Matematic˘ a

22

Problems

110. [ Gabriel Dospinescu ] Let a1 , a2 , . . . , an be real numbers and let S be a non-empty subset of {1, 2, . . . , n}. Prove that !2 X

ai

i∈S

≤

X

(ai + . . . + aj )2 .

1≤i≤j≤n

TST 2004, Romania 111. [ Dung Tran Nam ] Let x1 , x2 . . . , x2004 be real numbers in the interval [−1, 1] such that x31 + x32 + . . . + x32004 = 0. Find the maximal value of the x1 + x2 + · · · + x2004 .

112. [ Gabriel Dospinescu, C˘alin Popa ] Prove that if n ≥ 2 and a1 , a2 , . . . , an are real numbers with product 1, then √ 2n a21 + a22 + · · · + a2n − n ≥ · n n − 1(a1 + a2 + · · · + an − n). n−1 113. [ Vasile Cˆırtoaje ] If a, b, c are positive real numbers, then r r r 2a 2b 2c + + ≤ 3. a+b b+c c+a Gazeta Matematic˘ a 114. Prove the following inequality for positive real numbers x, y, z   1 1 1 9 + + (xy + yz + zx) ≥ (x + y)2 (y + z)2 (z + x)2 4 Iran, 1996 115. Prove that for any x, y in the interval [0, 1], p p p √ 1 + x2 + 1 + y 2 + (1 − x)2 + (1 − y)2 ≥ (1 + 5)(1 − xy).

116. [ Suranyi ] Prove that for any positive real numbers a1 , a2 , . . . , an the following inequality holds (n−1)(an1 +an2 +· · ·+ann )+na1 a2 . . . an ≥ (a1 +a2 +· · ·+an )(an−1 +an−1 +· · ·+an−1 ). n 1 2 Miklos Schweitzer Competition 117. Prove that for any x1 , x2 , . . . , xn > 0 with product 1, X 1≤i<j≤n

2

(xi − xj ) ≥

n X

x2i − n.

i=1

A generalization of Turkevici’s inequality

Old and New Inequalities

23

118. [ Gabriel Dospinescu ] Find the minimum value of the expression n r X a1 a2 . . . an 1 − (n − 1)ai i=1 where a1 , a2 , . . . , an <

1 add up to 1 and n > 2 is an integer. n−1

119. [ Vasile Cˆırtoaje ] Let a1 , a2 , . . . , an < 1 be nonnegative real numbers such that r √ a21 + a22 + . . . + a2n 3 ≥ . a= n 3 Prove that a1 a2 an na + + ... + ≥ . 2 2 2 1 − a1 1 − a2 1 − an 1 − a2 120. [ Vasile Cˆırtoaje, Mircea Lascu ] Let a, b, c, x, y, z be positive real numbers such that (a + b + c)(x + y + z) = (a2 + b2 + c2 )(x2 + y 2 + z 2 ) = 4. Prove that abcxyz <

1 . 36

121. [ Gabriel Dospinescu ] For a given n > 2, find the minimal value of the constant kn , such that if x1 , x2 , . . . , xn > 0 have product 1, then 1 1 1 √ +√ + ··· + √ ≤ n − 1. 1 + kn x1 1 + kn x2 1 + kn xn Mathlinks Contest 122. [ Vasile Cˆırtoaje, Gabriel Dospinescu ] For a given n > 2, find the maximal value of the constant kn such that for any x1 , x2 , . . . , xn > 0 for which x21 + x22 + · · · + x2n = 1 we have the inequality (1 − x1 )(1 − x2 ) . . . (1 − xn ) ≥ kn x1 x2 . . . xn .

CHAPTER 2

Solutions

25

26

Solutions

1. Prove that the inequality √ p p 3 2 b2 + (1 − c)2 + c2 + (1 − a)2 ≥ 2 holds for arbitrary real numbers a, b, c. p

a2 + (1 − b)2 +

K¨ omal First solution: Applying Minkowsky’s Inequality to the left-hand side we have p p p p a2 + (1 − b)2 + b2 + (1 − c)2 + c2 + (1 − a)2 ≥ (a + b + c)2 + (3 − a − b − c)2 . Denoting a + b + c = x we get 

3 (a + b + c) + (3 − a − b − c) = 2 x − 2 2

2

2 +

9 9 ≥ , 2 2

and the conclusion follows. Second solution: We have the inequalities p p p a2 + (1 − b)2 + b2 + (1 − c)2 + c2 + (1 − a)2 ≥ ≥

|a| + |1 − b| |b| + |1 − c| |c| + |1 − a| √ √ √ + + 2 2 2

√ 3 2 and because |x| + |1 − x| ≥ 1 for all real numbers x the last quantity is at least . 2 2. [ Dinu S ¸ erb˘ anescu ] If a, b, c ∈ (0, 1) prove that p √ abc + (1 − a)(1 − b)(1 − c) < 1. Junior TST 2002, Romania First solution: 1 1 Observe that x 2 < x 3 for x ∈ (0, 1). Thus √ √ 3 abc < abc, and p p (1 − a)(1 − b)(1 − c) < 3 (1 − a)(1 − b)(1 − c). By the AM-GM Inequality, √ abc < and p

(1 − a)(1 − b)(1 − c) <

p 3

√ 3

abc ≤

a+b+c , 3

(1 − a)(1 − b)(1 − c) ≤

(1 − a) + (1 − b) + (1 − c) . 3

Old and New Inequalities

27

Summing up, we obtain p √ a+b+c+1−a+1−b+1−c abc + (1 − a)(1 − b)(1 − c) < = 1, 3 as desired. Second solution: We have p √ √ √ √ √ abc + (1 − a)(1 − b)(1 − c) < b · c + 1 − b · 1 − c ≤ 1, by the Cauchy-Schwarz Inequality. Third solution:  π . The inequality Let a = sin2 x, b = sin2 y, c = sin2 z, where x, y, z ∈ 0, 2 becomes sin x · sin y · sin z + cos x · cos y · cos z < 1 and it follows from the inequalities sin x · sin y · sin z + cos x · cos y · cos z < sin x · sin y + cos x · cos y = cos(x − y) ≤ 1 3. [ Mircea Lascu ] Let a, b, c be positive real numbers such that abc = 1. Prove that √ √ b+c c+a a+b √ √ + √ + √ ≥ a + b + c + 3. a c b Gazeta Matematic˘ a Solution: From the AM-GM Inequality, we have r ! r r b+c c+a a+b bc ca ab √ + √ + √ ≥2 + + = a b c a c b r ! r r ! r ! r bc ca ca ab ab bc = + + + + + ≥ a b b c c a √ √ √ √ √ √ √ √ √ √ 6 ≥ 2( a + b + c) ≥ a + b + c + 3 abc = a + b + c + 3. r

4. If the equation x4 + ax3 + 2x2 + bx + 1 = 0 has at least one real root, then a + b2 ≥ 8. Tournament of the Towns, 1993 2

28

Solutions

Solution: Let x be the real root of the equation. Using the Cauchy-Schwarz Inequality we infer that
2 x4 + 2x2 + 1 2 2 a +b ≥ ≥ 8, x2 + x6 because the last inequality is equivalent to (x2 − 1)4 ≥ 0. 5. Find the maximum value of the expression x3 + y 3 + z 3 − 3xyz where x2 + y 2 + z 2 = 1 and x, y, z are real numbers. Solution: Let t = xy + yz + zx. Let us observe that (x3 + y 3 + z 3 − 3xyz)2 = (x + y + z)2 (1 − xy − yz − zx)2 = (1 + 2t)(1 − t)2 . 1 ≤ t ≤ 1. Thus, we must find the maximum value of the expression 2 1 (1 + 2t)(1 − t)2 , where − ≤ t ≤ 1. In this case we clearly have (1 + 2t)(t − 1)2 ≤ 1 ⇔ 2 t2 (3−2t) ≥ 0 and thus |x3 +y 3 +z 3 −3xyz| ≤ 1. We have equality for x = 1, y = z = 0 and thus the maximum value is 1. We also have −

6. Let a, b, c, x, y, z be positive real numbers such that x + y + z = 1. Prove that p ax + by + cz + 2 (xy + yz + zx)(ab + bc + ca) ≤ a + b + c. Ukraine, 2001 First solution: We will use the Cauchy-Schwarz Inequality twice. First, we can write ax+by+ p √ cz ≤ a2 + b2 + c2 · x2 + y 2 + z 2 and then we apply again the Cauchy-Schwarz Inequality to obtain: p ax + by + cz + 2 (xy + yz + zx)(ab + bc + ca) ≤ qX qX q X q X ≤ a2 · x2 + 2 ab · 2 xy ≤ qX qX X X X ≤ x2 + 2 xy · a2 + 2 ab = a. Second solution: The inequality being homogeneous in a, b, c we can assume that a + b + c = 1. We apply this time the AM-GM Inequality and we find that p ax+by+cz+2 (xy + yz + zx)(ab + bc + ca) ≤ ax+by+cz+xy+yz+zx+ab+bc+ca. Consequently, xy + yz + zx + ab + bc + ca =

1 − x2 − y 2 − z 2 1 − a2 − b2 − c2 + ≤ 1 − ax − by − cz, 2 2

Old and New Inequalities

29

the last one being equivalent to (x − a)2 + (y − b)2 + (z − c)2 ≥ 0. 7. [ Darij Grinberg ] If a, b, c are three positive real numbers, then a 2

(b + c)

+

b 2

(c + a)

+

c 2

(a + b)

≥

9 . 4 (a + b + c)

First solution: We rewrite the inequality as (a + b + c)

a 2

(b + c)

+

b 2

(c + a)

+

!

c 2

(a + b)

≥

9 . 4

Applying the Cauchy-Schwarz Inequality we get !  2 a b c a b c (a + b + c) ≥ + + . 2 + 2 + 2 b+c c+a a+b (b + c) (c + a) (a + b) It remains to prove that a b c 3 + + ≥ b+c c+a a+b 2 which is well-known. Second solution: Without loss of generality we may assume that a + b + c = 1. Now, consider x . A short computation of derivatives the function f : (0, 1) → (0, ∞), f (x) = (1 − x)2 shows that f is convex. Thus, we may apply Jensen’s Inequality and the conclusion follows. 8. [ Hojoo Lee ] Let a, b, c ≥ 0. Prove that p p p a4 + a2 b2 + b4 + b4 + b2 c2 + c4 + c4 + c2 a2 + a4 ≥ p p p ≥ a 2a2 + bc + b 2b2 + ca + c 2c2 + ab. Gazeta Matematic˘ a Solution: 4 2 2 4 4 2 2 4 We start from (a2 −b2 )2 ≥ 0. We √ rewrite it as 4a +4a b +4b ≥ 3a +6a b +3b . √ 3 2 It follows that a4 + a2 b2 + b2 ≥ (a + b2 ). 2 Using this observation, we find that X p 2 X 2 a4 + a2 b2 + b4 ≥ 3 a2 . But using the Cauchy-Schwarz Inequality we obtain  2  X  X X 2 X p
 a2 2a2 + bc ≤ 3 a2 a 2a2 + bc ≤ and the inequality is proved.

30

Solutions

9. If a, b, c are positive real numbers such that abc = 2, then √ √ √ a3 + b3 + c3 ≥ a b + c + b c + a + c a + b. When does equality hold? JBMO 2002 Shortlist First solution: Applying the Cauchy-Schwarz Inequality gives 3(a2 + b2 + c2 ) ≥ 3(a + b + c)2

(1)

and (a2 + b2 + c2 )2 ≤ (a + b + c)(a3 + b3 + c3 ).

(2)

These two inequalities combined yield (a2 + b2 + c2 )(a + b + c) 3 (a2 + b2 + c2 )[(b + c) + (a + c) + (a + b)] = 6 √ √ √ (a b + c + b a + c + c a + b)2 ≥ (3) 6 Using the AM-GM Inequality we obtain r p  √ √ √ a b + c + b a + c + c a + b ≥ 3 3 abc (a + b)(b + c)(c + a) q √ √ 3 3 ≥ 3 abc 8abc = 3 8 = 6. a3 + b3 + c3

≥

Thus √ √ √ √ √ √ (a b + c + b a + c + c a + b)2 ≥ 6(a b + c + b a + c + c a + b).

(4)

The desired inequality follows from (3) and (4). Second solution: We have p √ √ √ a b + c + b c + a + c a + b ≤ 2(a2 + b2 + c2 )(a + b + c). Using Chebyshev Inequality, we infer that p p 2(a2 + b2 + c2 )(a + b + c) ≤ 6(a3 + b3 + c3 ) and so it is enough to prove that a3 + b3 + c3 ≥ 3abc, which is true by the AM-GM √ Inequality. We have equality if a = b = c = 3 2. 10. [ Ioan Tomescu ] Let x, y, z > 0. Prove that xyz 1 ≤ 4. (1 + 3x)(x + 8y)(y + 9z)(z + 6) 7

Old and New Inequalities

31

When do we have equality? Gazeta Matematic˘ a Solution: First, we write the inequality in the following form     8y 9z 6 (1 + 3x) 1 + 1+ 1+ ≥ 74 . x y z But this follows immediately from Huygens Inequality. We have equality for 3 x = 2, y = , z = 1. 2 11. [ Mihai Piticari, Dan Popescu ] Prove that 5(a2 + b2 + c2 ) ≤ 6(a3 + b3 + c3 ) + 1, for all a, b, c > 0 with a + b + c = 1. Solution: Because a + b + c = 1, we have a3 + b3 + c3 = 3abc + a2 + b2 + c2 − ab − bc − ca. The inequality becomes 5(a2 + b2 + c2 )

≤

18abc + 6(a2 + b2 + c2 ) − 6(ab + bc + ca) + 1 ⇔

⇔ 18abc + 1 − 2(ab + bc + ca) + 1 ≥ 6(ab + bc + ca) ⇔ ⇔ 8(ab + bc + ca) ≤ 2 + 18abc ⇔ 4(ab + bc + ca) ≤ 1 + 9abc ⇔ ⇔ (1 − 2a)(1 − 2b)(1 − 2c) ≤ abc ⇔ ⇔ (b + c − a)(c + a − b)(a + b − c) ≤ abc, which is equivalent to Schur’s Inequality. 12. [ Mircea Lascu ] Let x1 , x2 , . . . , xn ∈ R, n ≥ 2 and a > 0 such that  x1 + a2 2a 2 2 2 x2 + . . . + xn = a and x1 + x2 + . . . + xn ≤ . Prove that xi ∈ 0, , for all n−1 n i ∈ {1, 2, . . . , n}. Solution: Using the Cauchy-Schwarz Inequality, we get
(a − x1 )2 ≤ (n − 1) x22 + x23 + · · · + x2n ≤ (n − 1)



 a2 − x21 . n−1

Thus,   2a a2 − 2ax1 + x21 ≤ a2 − (n − 1)x21 ⇔ x1 x1 − ≤0 n and the conclusion follows.

32

Solutions

13. [ Adrian Zahariuc ] Prove that for any a, b, c ∈ (1, 2) the following inequality holds √ √ √ b a a c c b √ √ + √ √ + √ √ ≥ 1. 4b c − c a 4c a − a b 4a b − b c

Solution: The fact that a, b, c ∈ (1, 2) makes all denominators positive. Then √ √ √ √ a b a √ √ ≥ ⇔ b(a + b + c) ≥ a(4b c − c a) ⇔ a+b+c 4b c − c a √ ⇔ (a + b)(b + c) ≥ 4b ac, √ √ the last one coming from a + b ≥ 2 ab and b + c ≥ 2 bc. Writing the other two inequalities and adding them up give the desired result. 14. For positive real numbers a, b, c such that abc ≤ 1, prove that c a b + + ≥ a + b + c. b c a

First solution: If ab + bc + ca ≤ a + b + c, then the Cauchy-Schwarz Inequality solves the problem: (a + b + c)2 1 1 1 + + a b c a2 c + b2 a + c2 b a b c ≥ a + b + c. + + = ≥ b c a abc abc Otherwise, the same inequality gives a b c a2 c + b2 a + c2 b + + = ≥ b c a abc

(ab + bc + ca)2 a+b+c ≥a+b+c abc

(here we have used the fact that abc ≤ 1). Second solution: 1 Replacing a, b, c by ta, tb, tc with t = √ preserves the value of the quantity in 3 abc the left-hand side of the inequality and increases the value of the right-hand side and makes at · bt · ct = abct3 = 1. Hence we may assume without loss of generality that y z x abc = 1. Then there exist positive real numbers x, y, z such that a = , b = , c = . x y z The Rearrangement Inequality gives x3 + y 3 + z 3 ≥ x2 y + y 2 z + z 2 x.

Old and New Inequalities

Thus

33

a b c x3 + y 3 + z 3 x2 y + y 2 z + z 2 x + + = ≥ =a+b+c b c a xyz xyz

as desired. Third solution: Using the AM-GM Inequality, we deduce that r 2 2a b 3 a + ≥3 ≥ 3a. b c bc 2b c 2c a Similarly, + ≥ 3b and + ≥ 3c. Adding these three inequalities, the conclusion c a a b is immediate. Forth solution: r r r 4 4 4 9 ca 9 bc 9 ab , y = , z = . Consequently, a = xy 2 , b = zx2 , c = yz 2 , Let x = c2 b2 a2 and also xyz ≤ 1. Thus, using the Rearrangement Inequality, we find that X a X x2 X x2 X X X = ≥ xyz = x3 ≥ xy 2 = a. b yz yz 15. [ Vasile Cˆırtoaje, Mircea Lascu ] Let a, b, c, x, y, z be positive real numbers such that a + x ≥ b + y ≥ c + z and a + b + c = x + y + z. Prove that ay + bx ≥ ac + xz. Solution: We have ay+bx−ac−xz = a(y−c)+x(b−z) = a(a+b−x−z)+x(b−z) = a(a−x)+(a+x)(b−z) = =

1 1 1 1 (a − x)2 + (a2 − x2 ) + (a + x)(b − z) = (a − x)2 + (a + x)(a − x + 2b − 2z) = 2 2 2 2 1 1 = (a − x)2 + (a + x)(b − c + y − z) ≥ 0. 2 2 The equality occurs when a = x, b = z, c = y and 2x ≥ y + z.

16. [ Vasile Cˆırtoaje, Mircea Lascu ] Let a, b, c be positive real numbers so that abc = 1. Prove that 6 3 ≥ . 1+ a+b+c ab + ac + bc Junior TST 2003, Romania Solution: 1 1 1 We set x = , y = , z = and observe that xyz = 1. The inequality is a b c equivalent to 3 6 1+ ≥ . xy + yz + zx x+y+z

34

Solutions

From (x + y + z)2 ≥ 3(xy + yz + zx) we get 1+

3 9 , ≥1+ xy + yz + zx (x + y + z)2

so it suffices to prove that 9 6 ≥ . (x + y + z)2 x+y+z 2  3 ≥ 0 and this ends the proof. The last inequality is equivalent to 1 − x+y+z 1+

17. Let a, b, c be positive real numbers. Prove that a3 b3 c3 a2 b2 c2 + 2+ 2 ≥ + + . 2 b c a b c a JBMO 2002 Shortlist First solution: We have a2 a3 ≥ + a − b ⇔ a3 + b3 ≥ ab(a + b) ⇔ (a − b)2 (a + b) ≥ 0, b2 b which is clearly true. Writing the analogous inequalities and adding them up gives a3 b3 c3 a2 b2 c2 a2 b2 c2 + 2+ 2 ≥ +a−b+ +b−c+ +c−a= + + . 2 b c a b c a b c a Second solution: By the Cauchy-Schwarz Inequality we have   2 2  3 b3 c3 a b2 c2 a + + ≥ + + , (a + b + c) b2 c2 a2 b c a so we only have to prove that a2 b2 c2 + + ≥ a + b + c. b c a But this follows immediately from the Cauchy-Schwarz Inequality. 18. Prove that if n > 3 and x1 , x2 , . . . , xn > 0 have product 1, then 1 1 1 + + ··· + > 1. 1 + x1 + x1 x2 1 + x2 + x2 x3 1 + xn + xn x1 Russia, 2004 Solution: a2 a3 a1 We use a similar form of the classical substitution x1 = , x2 = , . . . , xn = . a1 a2 an In this case the inequality becomes a1 a2 an + + ··· + >1 a1 + a2 + a3 a2 + a3 + a4 an + a1 + a2

Old and New Inequalities

35

and it is clear, because n > 3 and ai + ai+1 + ai+2 < a1 + a2 + · · · + an for all i. 19. [ Marian Tetiva ] Let x, y, z be positive real numbers satisfying the condition x2 + y 2 + z 2 + 2xyz = 1. Prove that 1 a) xyz ≤ ; 8 3 b) x + y + z ≤ ; 2 3 c) xy + xz + yz ≤ ≤ x2 + y 2 + z 2 ; 4 1 d) xy + xz + yz ≤ + 2xyz. 2 Solution: a) This is very simple. From the AM-GM Inequality, we have p 1 1 1 = x2 + y 2 + z 2 + 2xyz ≥ 4 4 2x3 y 3 z 3 ⇒ x3 y 3 z 3 ≤ ⇒ xyz ≤ . 2 · 44 8 b) Clearly, we must have x, y, z ∈ (0, 1). If we put s = x+y+z, we get immediately from the given relation s2 − 2s + 1 = 2 (1 − x) (1 − y) (1 − z) . Then, again by the AM-GM Inequality (1 − x, 1 − y, 1 − z being positive), we obtain 3  3  1−x+1−y+1−z 3−s 2 =2 . s − 2s + 1 ≤ 2 3 3 After some easy calculations this yields 2

2s3 + 9s2 − 27 ≤ 0 ⇔ (2s − 3) (s + 3) ≤ 0 and the conclusion is plain. c) These inequalities are simple consequences of a) and b): 2

xy + xz + yz ≤

(x + y + z) 1 9 3 ≤ · = 3 3 4 4

and 1 3 = . 8 4 d) This is more delicate; we first notice that there are always two of the three 1 numbers, both greater (or both less) than . Because of symmetry, we may assume 2 1 1 that x, y ≤ , or x, y ≥ and then 2 2 1 (2x − 1) (2y − 1) ≥ 0 ⇔ x + y − 2xy ≤ . 2 x2 + y 2 + z 2 = 1 − 2xyz ≥ 1 − 2 ·

36

Solutions

On the other hand, 1

=

x2 + y 2 + z 2 + 2xyz ≥ 2xy + z 2 + 2xyz ⇒

⇒ 2xy (1 + z) ≤ 1 − z 2 ⇒ 2xy ≤ 1 − z. Now, we only have to multiply side by side the inequalities from above x + y − 2xy ≤

1 2

and z ≤ 1 − 2xy to get the desired result: 1 1 − xy ⇔ xy + xz + yz ≤ + 2xyz. 2 2 It is not important in the proof, but also notice that   1 1 x + y − 2xy = xy + − 2 > 0, x y xz + yz − 2xyz ≤

because

1 1 and are both greater than 1. x y

Remark. 1) One can obtain some other inequalities, using z + 2xy ≤ 1 and the two likes y + 2xz ≤ 1, x + 2yz ≤ 1. For example, multiplying these inequalities by z, y, x respectively and adding the new inequalities, we get x2 + y 2 + z 2 + 6xyz ≤ x + y + z, or 1 + 4xyz ≤ x + y + z. 2) If ABC is any triangle, the numbers B C A , y = sin , z = sin 2 2 2 satisfy the condition of this problem; conversely, if x, y, z > 0 verify x = sin

x2 + y 2 + z 2 + 2xyz = 1 then there is a triangle ABC so that A B C , y = sin , z = sin . 2 2 2 According to this, new proofs can be given for such inequalities. x = sin

Old and New Inequalities

37

20. [ Marius Olteanu ] Let x1 , x2 , x3 , x4 , x5 ∈ R so that x1 + x2 + x3 + x4 + x5 = 0. Prove that | cos x1 | + | cos x2 | + | cos x3 | + | cos x4 | + | cos x5 | ≥ 1. Gazeta Matematic˘ a Solution: It is immediate to prove that | sin(x + y)| ≤ min{| cos x| + | cos y|, | sin x| + | sin y|} and | cos(x + y)| ≤ min{| sin x| + | cos y|, | sin y| + | cos x|}. Thus, we infer that ! ! 5 5 X X 1 = | cos xi | ≤ | cos x1 |+| sin xi | ≤ | cos x1 |+| cos x2 |+| cos(x3 +x4 +x5 )|. i=1

i=2

But in the same manner we can prove that | cos(x3 + x4 + x5 )| ≤ | cos x3 | + | cos x4 | + | cos x5 | and the conclusion follows. 21. [ Florina Cˆ arlan, Marian Tetiva ] Prove that if x, y, z > 0 satisfy the condition

then xy + xz + yz ≥ 3 +

√

x + y + z = xyz p √ x2 + 1 + y 2 + 1 + z 2 + 1.

First solution: We have √ 2 √ √ xyz = x + y + z ≥ 2 xy + z ⇒ z ( xy) − 2 xy − z ≥ 0. Because the positive root of the trinomial zt2 − 2t − z is √ 1 + 1 + z2 , z we get from here √ p 1 + 1 + z2 √ √ xy ≥ ⇔ z xy ≥ 1 + 1 + z 2 . z Of course, we have two other similar inequalities. Then, √ √ √ xy + xz + yz ≥ x yz + y xz + z xy ≥ p p p ≥ 3 + x2 + 1 + y 2 + 1 + z 2 + 1, and we have both a proof of the given inequality, and a little improvement of it.

38

Solutions

Second solution: Another improvement is as follows. Start from 1 1 1 1 1 1 + 2+ 2 ≥ + + = 1 ⇒ x2 y 2 + x2 z 2 + y 2 z 2 ≥ x2 y 2 z 2 , 2 x y z xy xz yz which is equivalent to 2

2

(xy + xz + yz) ≥ 2xyz (x + y + z) + x2 y 2 z 2 = 3 (x + y + z) . Further on, 2

2

(xy + xz + yz − 3) = (xy + xz + yz) − 6 (xy + xz + yz) + 9 ≥
2 ≥ 3 (x + y + z) − 6 (xy + xz + yz) + 9 = 3 x2 + y 2 + z 2 + 9, so that xy + xz + yz ≥ 3 +

p 3 (x2 + y 2 + z 2 ) + 9.

But p p p p 3 (x2 + y 2 + z 2 ) + 9 ≥ x2 + 1 + y 2 + 1 + z 2 + 1 is a consequence of the Cauchy-Schwarz Inequality and we have a second improvement and proof for the desired inequality: p xy + xz + yz ≥ 3 + 3 (x2 + y 2 + z 2 ) + 9 ≥ p p p ≥ 3 + x2 + 1 + y 2 + 1 + z 2 + 1.

22. [ Laurent¸iu Panaitopol ] Prove that 1 + x2 1 + y2 1 + z2 + + ≥ 2, 1 + y + z2 1 + z + x2 1 + x + y2 for any real numbers x, y, z > −1. JBMO, 2003 Solution: Let us observe that y ≤

1 + y2 and 1 + y + z 2 > 0, so 2

1 + x2 ≥ 1 + y + z2

1 + x2 1 + y2 1 + z2 + 2

and the similar relations. Setting a = 1 + x2 , b = 1 + y 2 , c = 1 + z 2 , it is sufficient to prove that a b c + + ≥1 (1) 2c + b 2a + c 2b + a

Old and New Inequalities

39

C + 4B − 2A for any a, b, c > 0. Let A = 2c + b, B = 2a + c, C = 2b + a. Then a = , 9 A + 4C − 2B B + 4A − 2C ,c= and the inequality (1) is rewritten as b= 9 9   A B B C A C + + +4 + + ≥ 15. A B C A B C Because A, B, C > 0, we have from the AM-GM Inequality that r A B B C C 3 A + + ≥3 · · =3 A B C B C A B C A and + + ≥ 3 and the conclusion follows. A B C An alternative solution for (1) is by using the Cauchy-Schwarz Inequality: a b c a2 b2 c2 (a + b + c)2 + + = + + ≥ ≥ 1. 2c + b 2a + c 2b + a 2ac + ab 2ab + cb 2bc + ac 3(ab + bc + ca) 23. Let a, b, c > 0 with a + b + c = 1. Show that a2 + b b2 + c c2 + a + + ≥ 2. b+c c+a a+b

Solution: Using the Cauchy-Schwarz Inequality, we find that X 2 a2 + 1 X a2 + b X X . ≥X b+c a2 (b + c) + a2 + ab And so it is enough to prove that 2 X X 2 a2 + 1 X X X X X a2 (b + c) + 2 ab. ≥2 ⇔ 1+ a2 ≥ 2 a2 (b + c) + a2 + ab The last inequality can be transformed as follows X 2 X 2 X X 1+ a2 ≥ 2 a2 (b + c) + 2 ab ⇔ 1 + a2 ≥  X 2 X X X X X ≥ 2 a2 − 2 a3 + 2 ab ⇔ a2 + 2 a3 ≥ a2 , and it is true, because X X

a3 ≥

3

a2 (Chebyshev’s Inequality)

and X

2

a

2

X ≥

3

a2 .

40

Solutions

24. Let a, b, c ≥ 0 such that a4 + b4 + c4 ≤ 2(a2 b2 + b2 c2 + c2 a2 ). Prove that a2 + b2 + c2 ≤ 2(ab + bc + ca). Kvant, 1988 Solution: The condition X

a4 ≤ 2

X

a2 b2

is equivalent to (a + b + c)(a + b − c)(b + c − a)(c + a − b) ≥ 0. In any of the cases a = b + c, b = c + a, c = a + b, the inequality X X a2 ≤ 2 ab is clear. So, suppose a + b 6= c, b + c 6= a, c + a 6= b. Because at most one of the numbers b + c − a, c + a − b, a + b − c is negative and their product is nonnegative, all of them are positive. Thus, we may assume that a2 < ab + ac, b2 < bc + ba, c2 < ca + cb and the conclusion follows. 25. Let n ≥ 2 and x1 , ..., xn be positive real numbers satisfying 1 1 1 1 + + ... + = . x1 + 1998 x2 + 1998 xn + 1998 1998 Prove that

√ n

x1 x2 ...xn ≥ 1998. n−1 Vietnam, 1998

Solution: 1998 Let = ai . The problem reduces to proving that for any positive real 1998 + xi numbers a1 , a2 , . . . , an such that a1 + a2 + · · · + an = 1 we have the inequality  n  Y 1 − 1 ≥ (n − 1)n . ai i=1 This inequality can be obtained by multiplying the inequalities 1 −1 ai

a1 + · · · + ai−1 + ai+1 + · · · + an ≥ ai r a1 . . . ai−1 ai+1 . . . an ≥ (n − 1) n−1 an−1 i =

Old and New Inequalities

26. [ Marian Tetiva ] Consider positive real numbers x, y, z so that x2 + y 2 + z 2 = xyz. Prove the following inequalities a) xyz ≥ 27; b) xy + xz + yz ≥ 27; c) x + y + z ≥ 9; d) xy + xz + yz ≥ 2 (x + y + z) + 9. Solution: a), b), c) Using well-known inequalities, we have p 3 2 xyz = x2 + y 2 + z 2 ≥ 3 3 x2 y 2 z 2 ⇒ (xyz) ≥ 27 (xyz) , which yields xyz ≥ 27. Then xy + xz + yz ≥ 3 and

q √ 3 3 2 (xyz) ≥ 3 272 = 27

√ √ 3 x + y + z ≥ 3 3 xyz ≥ 3 27 = 9. d) We notice that x2 < xyz ⇒ x < yz and the likes. Consequently, xy < yz · xz ⇒ 1 < z 2 ⇒ z > 1.

Hence all the three numbers must be greater than 1. Set a = x − 1, b = y − 1, c = z − 1. We then have a > 0, b > 0, c > 0 and x = a + 1, y = b + 1, z = c + 1. Replacing these in the given condition we get 2

2

2

(a + 1) + (b + 1) + (c + 1) = (a + 1) (b + 1) (c + 1) which is the same as a2 + b2 + c2 + a + b + c + 2 = abc + ab + ac + bc. If we put q = ab + ac + bc, we have q ≤ a2 + b2 + c2 ,

p 3q ≤ a + b + c

and abc ≤

 q  32 3

3

(3q) 2 = . 27

41

42

Solutions

Thus p 3q + 2 ≤ a2 + b2 + c2 + a + b + c + 2 = √
3 3q = abc + ab + ac + bc ≤ +q 27 √ and setting x = 3q, we have q+

x+2≤

x3 2 ⇔ (x − 6) (x + 3) ≥ 0. 27

Finally, p 3q = x ≥ 6 ⇒ q = ab + ac + bc ≥ 12. Now, recall that a = x − 1, b = y − 1, c = z − 1, so we get (x − 1) (y − 1)

+

(x − 1) (z − 1) + (y − 1) (z − 1) ≥ 12 ⇒

⇒ xy + xz + yz ≥ 2 (x + y + z) + 9; and we are done. One can also prove the stronger inequality xy + xz + yz ≥ 4 (x + y + z) − 9. Try it! 27. Let x, y, z be positive real numbers with sum 3. Prove that √ √ √ x + y + z ≥ xy + yz + zx. Russia, 2002 Solution: Rewrite the inequality in the form √ √ √ x2 + 2 x + y 2 + 2 y + z 2 + 2 z ≥ x2 + y 2 + z 2 + 2xy + 2yz + 2zx ⇔ √ √ √ ⇔ x2 + 2 x + y 2 + 2 y + z 2 + 2 z ≥ 9. Now, from the AM-GM Inequality, we have √ √ √ √ 3 x2 + 2 x = x2 + x + x ≥ 3 x2 · x = 3x, √ √ y 2 + 2 y ≥ 3y, z 2 + 2 z ≥ 3z, hence

√ √  √ x2 + y 2 + z 2 + 2 x + y + z ≥ 3(x + y + z) ≥ 9.

28. [ D. Olteanu ] Let a, b, c be positive real numbers. Prove that a+b a b+c b c+a c 3 · + · + · ≥ . b + c 2a + b + c c + a 2b + c + a a + b 2c + a + b 4 Gazeta Matematic˘ a

Old and New Inequalities

43

Solution: We set x = a + b, y = b + c and z = a + c and after a few computations we obtain the equivalent form x y z x y z 9 + + + + + ≥ . y z x x+y y+z z+x 2 But using the Cauchy-Schwarz Inequality, x y z x y z (x + y + z)2 (x + y + z)2 = + + + + + ≥ + y z x x+y y+z z+x xy + yz + zx xy + yz + zx + x2 + y 2 + z 2 2(x + y + z)4 8(x + y + z)4 ≥ 2 2 2 2 2(xy + yz + zx)(xy + yz + zx + x + y + z ) (xy + yz + zx + (x + y + z)2 ) and the conclusion follows. 29. For any positive real numbers a, b, c show that the following inequality holds a b c c+a a+b b+c + + ≥ + + . b c a c+b a+c b+a India, 2002 Solution: a b c Let us take = x, = y, = z. Observe that b c a 1 + xy 1−x a+c = =x+ . b+c 1+y 1+y Using similar relations, the problem reduces to proving that if xyz = 1, then x−1 y−1 z−1 + + ≥0 ⇔ y+1 z+1 x+1 ⇔ (x2 − 1)(z + 1) + (y 2 − 1)(x + 1) + (z 2 − 1)(y + 1) ≥ 0 ⇔ X X X ⇔ x2 z + x2 ≥ x + 3. But Inequality we have X this inequality is very easy. Indeed, using Xthe AM-GM X 2 2 x z ≥ 3 and so it remains to prove that x ≥ x, which follows from the inequalities X 2 x X X x2 ≥ ≥ x. 3 30. Let a, b, c be positive real numbers. Prove that a3 b3 c3 3(ab + bc + ca) + + ≥ . 2 2 2 2 2 2 b − bc + c c − ac + a a − ab + b a+b+c Proposed for the Balkan Mathematical Olympiad

44

Solutions

First solution: Since a + b + c ≥

b2

3(ab + bc + ca) , it suffices to prove that: a+b+c

b3 c3 a3 + 2 + 2 ≥ a + b + c. 2 2 − bc + c c − ca + a a − ab + b2

From the Cauchy-Schwartz Inequality, we get P 2
2 X a a3 ≥P . b2 − bc + c2 a(b2 − bc + c2 ) Thus we have to show that (a2 + b2 + c2 )2 ≥ (a + b + c) ·

X

a(b2 − bc + c2 ).

This inequality is equivalent to a4 + b4 + c4 + abc(a + b + c) ≥ ab(a2 + b2 ) + bc(b2 + c2 ) + ca(c2 + a2 ), which is just Schur’s Inequality. Remark. The inequality b2

b3 c3 a3 + 2 + 2 ≥a+b+c 2 2 − bc + c c − ca + a a − ab + b2

was proposed by Vasile Cˆ artoaje in Gazeta Matematic˘ a as a special case (n = 3) of the more general inequality 2bn − cn − an 2cn − an − bn 2an − bn − cn + 2 + 2 ≥ 0. 2 2 2 b − bc + c c − ca + a a − ab + b2 Second solution: Rewrite our inequality as X (b + c)a3 b3 + c3

≥

3(ab + bc + ca) . a+b+c

But this follows from a more general result: If a, b, c, x, y, z > 0 then X a(y + z) b+c

X

xy ≥3 X . x

But this inequality is an immediate (and weaker) consequence of the result from problem 101. Third solution: Let A=

X b2

a3 − bc + c2

Old and New Inequalities

and B=

X b2

45

X (b + c)(b2 − bc + c2 ) X b3 + c3 = =2 a. 2 2 2 − bc + c b − bc + c

So we get A+B

= =

X

a3

 X

1 b2 − bc + c2



=   X 2 1 X 1 X √ 1 ≥ (b + c)(b2 − bc + c2 ) b+c , 2 2 2 b − bc + c 2

from the Cauchy-Schwarz Inequality. Hence 2 X X√ X √ 1 X √ b+c −2 a= a+b· b+c− a. A≥ 2 We denote X ab √ √ Aa = c + a · b + a − a = qX ab + a2 + a X 2 a ≥ and the similar relations with Ab and Ac . So A ≥ Aa +Ab +Ac . But because X  3 ab we get v  u X 2 X X u a ab 3 ab t  Aa ≥ s X  = X 2  + a2 − a   2 3 a a + a2 + a 3 and also the similar inequalities are true. So we only need to prove that v  u X  2 s u 2 a X t X 1   a 2 − a ≥ a + b + c ⇔  + a + ≥ 2.   3 3 a+b+c r We consider the convex function f (t) =

1 + t2 . Using Jensen’s Inequality 3

we finally deduce that s

 2 1 a + ≥ 2. 3 a+b+c We have equality if and only if a = b = c. X

31. [ Adrian Zahariuc ] Consider the pairwise distinct integers x1 , x2 , . . . , xn , n ≥ 0. Prove that x21 + x22 + · · · + x2n ≥ x1 x2 + x2 x3 + · · · + xn x1 + 2n − 3.

46

Solutions

Solution: The inequality can be rewritten as n X

(xi − xi+1 )2 ≥ 2(2n − 3).

i=1

Let xm = min{x1 , x2 , . . . , xn } and xM = max{x1 , x2 , . . . , xn }. Suppose without loss of generality that m < M . Let S1 = (xm − xm+1 )2 + · · · + (xM −1 − xM )2 and S2 = (xM − xM +1 )2 + · · · + (xn − x1 )2 + (x1 − x2 )2 + · · · + (xm−1 − xm )2 . !2 k k X 1 X 2 The inequality ai ≥ ai (which follows from the Cauchy-Schwarz Ink i=1 i=1 equality) implies that (xM − xm )2 S1 ≥ M −m and (xM − xm )2 S2 ≥ . n − (M − m) So   n X 1 1 (xi − xi+1 )2 = S1 + S2 ≥ (xM − xm )2 + ≥ M − m n − (M − m) i=1 ≥ (n − 1)2 But

n X

4 4 = 4n − 8 + > 4n − 8. n n

(xi − xi+1 )2 ≡

i=1

so

n X

xi − xi+1 = 0 mod 2

i=1 n X

(xi − xi+1 )2 ≥ 4n − 6

i=1

and the problem is solved. 32. [ Murray Klamkin ] Find the maximum value of the expression x21 x2 + x22 x3 + · · · + x2n−1 xn + x2n x1 when x1 , x2 , . . . , xn−1 , xn ≥ 0 add up to 1 and n > 2. Crux Mathematicorum Solution: 4 First of all, it is clear that the maximum is at least , because this value is 27 2 1 attained for x1 = , x2 = , x3 = · · · = xn = 0. Now, we will prove by induction that 3 3 4 x21 x2 + x22 x3 + · · · + x2n−1 xn + x2n x1 ≤ 27

Old and New Inequalities

47

for all x1 , x2 , . . . , xn−1 , xn ≥ 0 which add up to 1. Let us prove first the inductive step. Suppose the inequality is true for n and we will prove it for n + 1. We can of course assume that x2 = min{x1 , x2 , . . . , xn+1 }. But this implies that x21 x2 + x22 x3 + · · · + x2n x1 ≤ (x1 + x2 )2 x3 + x23 x4 + · · · + x2n−1 xn + x2n (x1 + x2 ). But from the inductive hypothesis we have (x1 + x2 )2 x3 + x23 x4 + · · · + x2n−1 xn + x2n (x1 + x2 ) ≤

4 27

4 and the inductive step is proved. Thus, it remains to prove that a2 b + b2 c + c2 a ≤ 27 if a + b + c = 1. We may of course assume that a is the greatest among a, b, c. In this  c 2  c case the inequality a2 b + b2 c + c2 a ≤ a + follows immediately from · b+ 2 2 2 2 a c ac abc ≥ b2 c, ≥ . Because 2 2 v u c c c c 2 u 3 a+ a+ b + a + t 2 + 2 +b+ c ≥3 2 2 , 1= 2 2 2 4 4 we have proved that a2 b + b2 c + c2 a ≤ and this shows that the maximum is indeed 27 4 . 27 33. Find the maximum value of the constant c such that for any x1 , x2 , . . . , xn , · · · > 0 for which xk+1 ≥ x1 + x2 + · · · + xk for any k, the inequality √ √ √ √ x1 + x2 + · · · + xn ≤ c x1 + x2 + · · · + xn also holds for any n. IMO Shortlist, 1986 Solution: First, let us see what happens if xk+1 and x1 +x2 +· · ·+xk are close for any k. For example, we can take xk = 2k , because in this case we have x1 +x2 +· · ·+xk = xk+1 −2. Thus, we find that n X √ k 2 k=1 c≥ v u n uX t 2k k=1

√ √ for any n. Taking the limit, we find that c ≥ 1 + 2. Now, let us prove that 1 + 2 works. We will prove the inequality √ √ √ √ √ x1 + x2 + · · · + xn ≤ (1 + 2) x1 + x2 + · · · + xn

48

Solutions

by induction. For n = 1 or n = 2 it is clear. Suppose it is true for n and we will √ √ √ √ √ √ prove that x1 + x2 + · · · + xn + xn+1 ≤ (1 + 2) x1 + x2 + · · · + xn + xn+1 . Of course, it is enough to prove that √ p
√ √ xn+1 ≤ (1 + 2) x1 + x2 + · · · + xn+1 − x1 + x2 + · · · + xn which is equivalent to √ √ p √ x1 + x2 + · · · + xn + x1 + x2 + · · · + xn+1 ≤ (1 + 2) xn+1 . But this one follows because x1 + x2 + · · · + xn ≤ xn+1 . 34. Given are positive real numbers a, b, c and x, y, z, for which a + x = b + y = c + z = 1. Prove that 1 1 1 + + ≥ 3. (abc + xyz) ay bz cx Russia, 2002 Solution: Let us observe that abc + xyz = (1 − b)(1 − c) + ac + ab − a. Thus, 1−c c abc + xyz + −1= . a 1−b a(1 − b) Using these identities we deduce immediately that   1 1 1 a b c 1−c 1−a 1−b 3 + (xyz + abc) + + = + + + + + . ay bz cx 1−c 1−a 1−b a b c Now, all we have to do is apply the AM-GM Inequality a b c 1−c 1−a 1−b + + + + + ≥ 6. 1−c 1−a 1−b a b c 35. [ Viorel Vˆ ajˆ aitu, Alexandru Zaharescu ] Let a, b, c be positive real numbers. Prove that ab bc ca 1 + + ≤ (a + b + c). a + b + 2c b + c + 2a c + a + 2b 4 Gazeta Matematic˘ a First solution: We have the following chain of inequalities  X X X ab  1 ab ab 1 a+b+c = ≤ + = . a + b + 2c a+c+b+c 4 a+c b+c 4

Old and New Inequalities

49

Second solution: Because the inequality is homogeneous we can consider without loss of generality that a + b + c = 1 and so the inequality is equivalent to X 1 1 ≤ . a(a + 1) 4abc 1 1 1 We have = − , so the inequality is equivalent to t(t + 1) t t+1 X1 X 1 1 ≤ + . a a + 1 4abc We will prove now the following intercalation: X1 X 1 9 1 1 ≤ + ≤ + . a 4 4abc a + 1 4abc The inequality in the right follows from the Cauchy-Schwarz Inequality:  X X  1 (a + 1) ≥ 9 a+1 X X and the identity (a + 1) = 4. The inequality in the left can be written as ab ≤ 1 + 9abc , which is exactly Schur’s Inequality. 4 36. Find the maximum value of the expression a3 (b + c + d) + b3 (c + d + a) + c3 (d + a + b) + d3 (a + b + c) where a, b, c, d are real numbers whose sum of squares is 1. Solution: The X idea is to observe that a3 (b+c+d)+b3 (c+d+a)+c3 (d+a+b)+d3 (a+b+c) is equal to ab(a2 +b2 ). Now, because the expression ab(a2 +b2 ) appears when writing 4 (a − b) , let us see how the initial expression can be written: X X a4 + b4 + 6a2 b2 − (a − b)4 ab(a2 + b2 ) = = 4 X X X P 3 a4 + 6 a2 b2 − (a − b)4 3 − (a − b)4 3 = = ≤ . 4 4 4 1 The maximum is attained for a = b = c = d = . 2 37. [ Walther Janous ] Let x, y, z be positive real numbers. Prove that x y z p p p + + ≤ 1. x + (x + y)(x + z) y + (y + z)(y + x) z + (z + x)(z + y) Crux Mathematicorum

50

Solutions

First solution: √ √ √ We have (x + y)(x + z) = xy + (x2 + yz) + xz ≥ xy + 2x yz + xz = ( xy + xz)2 . Hence X X x x p √ . ≤ √ x + xy + xz x + (x + y)(x + z) But

√ X x x √ = √ √ =1 √ √ x + xy + xz x+ y+ z and this solves the problem. X

Second solution: p √ From Huygens Inequality we have (x + y)(x + z) ≥ x + yz and using this inequality for the similar ones we get x y z p p p + + ≤ x + (x + y)(x + z) y + (y + z)(y + x) z + (z + x)(z + y) x y z √ + ≤ √ + √ . 2x + yz 2y + zx 2z + xy √ √ √ yz xy zx ,b= ,c= and the inequality becomes Now, we denote with a = x y z 1 1 1 + + ≤ 1. 2+a 2+b 2+c From the above notations we can see that abc = 1, so the last inequality becomes after clearing the denominators ab + bc + ca ≥ 3, which follows from the AM-GM Inequality. 38. Suppose that a1 < a2 < ... < an are real numbers for some integer n ≥ 2. Prove that a1 a42 + a2 a43 + ... + an a41 ≥ a2 a41 + a3 a42 + ... + a1 a4n . Iran, 1999 Solution: A quick look shows that as soon as we prove the inequality for n = 3, it will be proved by induction for larger n. Thus, we must prove that for any a < b < c we have ab(b3 − a3 ) + bc(c3 − b3 ) ≥ ca(c3 − a3 ) ⇔ (c3 − b3 )(ac − bc) ≤ (b3 − a3 )(ab − ac). Because a < b < c, the last inequality reduces to a(b2 + ab + a2 ) ≤ c(c2 + bc + b2 ). And this last inequality is equivalent to (c − a)(a2 + b2 + c2 + ab + bc + ca) ≥ 0, which is clear. 39. [ Mircea Lascu ] Let a, b, c be positive real numbers. Prove that   b+c c+a a+b a b c + + ≥4 + + . a b c b+c c+a a+b

Old and New Inequalities

Solution: Using the inequality

51

1 1 1 ≤ + we infer that x+y 4x 4y

a a 4b b b 4c c c 4a ≤ + , ≤ + and ≤ + . b+c b c a+c a c a+b a b Adding up these three inequalities, the conclusion follows. 40. Let a1 , a2 , . . . , an > 1 be positive integers. Prove that at least one of the √ √ √ √ √ numbers a1 a2 , a2 a3 , . . . , an−1 an , an a1 is less than or equal to 3 3. Adapted after a well-known problem Solution: 1 1 1 1 a1 > 3 3 for all i. First, we will prove that n n ≤ 3 3 for all Suppose we have ai+1 natural number n. For n = 1, 2, 3, 4 it is clear. Suppose the inequality is true for n > 3 and let us prove it for n + 1. This follows from the fact that √ √ n+1 n 1 1 n+1 3 3 1 + ≤ 1 + < 3 ⇒ 3 3 = 3 · 33 ≥ · n = n + 1. n 4 n 1 a

1 1

a

i+1 i > 3 3 ≥ ai+1 ⇒ ai+1 > ai for all i, Thus, using this observation, we find that ai+1 which means that a1 < a2 < · · · < an−1 < an < a1 , contradiction.

41. [ Mircea Lascu, Marian Tetiva ] Let x, y, z be positive real numbers which satisfy the condition xy + xz + yz + 2xyz = 1. Prove that the following inequalities hold 1 a) xyz ≤ ; 8 3 b) x + y + z ≥ ; 2 1 1 1 c) + + ≥ 4 (x + y + z); x y z 2 1 1 (2z − 1) 1 , where z = max {x, y, z}. d) + + − 4 (x + y + z) ≥ x y z (z (2z + 1)) Solution: a) We put t3 = xyz; according to the AM-GM Inequality we have 2

1 = xy + xz + yz + 2xyz ≥ 3t2 + 2t3 ⇔ (2t − 1) (t + 1) ≤ 0, 1 1 , this meaning that xyz ≤ . 2 8 b) Denote also s = x + y + z; the following inequalities are well-known

therefore 2t − 1 ≤ 0 ⇔ t ≤

2

(x + y + z) ≥ 3 (xy + xz + yz) and 3

(x + y + z) ≥ 27xyz;

52

Solutions

then we have 2s3 ≥ 54xyz = 27 − 27 (xy + xz + yz) ≥ 27 − 9s2 , i. e. 2

2s3 + 9s2 − 27 ≥ 0 ⇔ (2s − 3) (s + 3) ≥ 0, 3 where from 2s − 3 ≥ 0 ⇔ s ≥ . 2 1 Or, because p ≤ , we have 8   2 9 s2 ≥ 3q = 3 (1 − 2p) ≥ 3 1 − = ; 8 4 if we put q = xy + xz + yz, p = xyz. Now, one can see the following is also true q = xy + xz + yz ≥

3 . 4

1 c) The three numbers x, y, z cannot be all less than , because, in this case we 2 get the contradiction xy + xz + yz + 2xyz <

1 3 + 2 · = 1; 4 8

1 . 2 √ We have 1 = (2z + 1) xy + z (x + y) ≥ (2z + 1) xy + 2z xy, which can also

√ √ be written in the form (2z + 1) xy − 1 xy + 1 ≤ 0; and this one yields the inequality 1 xy ≤ 2. (2z + 1) because of symmetry we may assume then that z ≥

2

(x + y) + z (x + y), conse4 quently ((2z + 1) (x + y) − 2) (x + y + 2) ≥ 0, which shows us that We also have 1 = (2z + 1) xy + z (x + y) ≤ (2z + 1)

x+y ≥

2 . 2z + 1

The inequality to be proved 1 1 1 + + ≥ 4 (x + y + z) x y z can be also rearranged as   1 4z 2 − 1 (2z − 1) (2z + 1) (x + y) −4 ≥ = . xy z z From the above calculations we infer that    2 (2z − 1) (2z + 3) 1 2  2 (x + y) −4 ≥ (2z + 1) − 4 = xy 2z + 1 2z + 1

Old and New Inequalities

53

1 (the assumption z ≥ allows the multiplication of the inequalities side by side), and 2 this means that the problem would be solved if we proved 2z + 1 2 (2z + 3) ≥ ⇔ 4z 2 + 6z ≥ 4z 2 + 4z + 1; 2z + 1 z 1 but this follows for z ≥ and we are done. 2 1 d) Of course, if z is the greatest from the numbers x, y, z, then z ≥ ; we saw 2 that   1 1 1 2 + − 4 (x + y) = (x + y) −4 ≥ (2z + 1) x y xy 2z + 1 2

=

1 (2z − 1) 2 (2z − 1) (2z + 3) = 4z − + , 2z + 1 z z (2z + 1)

where from we get our last inequality 2

1 1 1 (2z − 1) + + − 4 (x + y + z) ≥ . x y z z (2z + 1) Of course, in the right-hand side z could be replaced by any of the three numbers 1 which is ≥ (two such numbers could be, surely there is one). 2 Remark. It is easy to see that the given condition implies the existence of positive numbers a b c a, b, c such that x = ,y = ,z = . And now a), b) and c) reduce b+c c+a a+b immediately to well-known inequalities! Try to prove using this substitution d). 42. [ Manlio Marangelli ] Prove that for any positive real numbers x, y, z, 3(x2 y + y 2 z + z 2 x)(xy 2 + yz 2 + zx2 ) ≥ xyz(x + y + z)3 .

Solution: Using the AM-GM Inequality, we find that √ 3y 3 xyz 1 y2 z xy 2 + + ≥ p 3 3 y 2 z + z 2 x + x2 y yz 2 + zx2 + xy 2 3 · (y 2 z + z 2 x + x2 y) · (yz 2 + zx2 + xy 2 ) and two other similar relations

√ 3z 3 xyz 1 z2x yz 2 + + ≥ p 3 3 y 2 z + z 2 x + x2 y yz 2 + zx2 + xy 2 3 · (y 2 z + z 2 x + x2 y) · (yz 2 + zx2 + xy 2 ) √ 3x 3 xyz 1 x2 y zx2 + + ≥ p 3 3 y 2 z + z 2 x + x2 y yz 2 + zx2 + xy 2 3 · (y 2 z + z 2 x + x2 y) · (yz 2 + zx2 + xy 2 )

Then, adding up the three relations, we find exactly the desired inequality.

54

Solutions

43. [ Gabriel Dospinescu ] Prove that if a, b, c are real numbers such that max{a, b, c} − min{a, b, c} ≤ 1, then 1 + a3 + b3 + c3 + 6abc ≥ 3a2 b + 3b2 c + 3c2 a Solution: Clearly, we may assume that a = min{a, b, c} and let us write b = a + x, c = a + y, where x, y ∈ [0, 1]. It is easy to see that a3 + b3 + c3 − 3abc = 3a(x2 − xy + y 2 ) + x3 + y 3 and a2 b + b2 c + c2 a − 3abc = a(x2 − xy + y 2 ) + x2 y. So, the inequality becomes 1 + x3 + y 3 ≥ 3x2 y. But this follows from the fact that 1 + x3 + y 3 ≥ 3xy ≥ 3x2 y, because 0 ≤ x, y ≤ 1. 44. [ Gabriel Dospinescu ] Prove that for any positive real numbers a, b, c we have       a2 b2 c2 1 1 1 27 + 2 + 2+ 2+ ≥ 6(a + b + c) + + . bc ca ab a b c Solution: By expanding the two sides, the inequality is equivalent to 2abc(a3 + b3 + c3 + 3abc − a2 b − a2 c − b2 a − b2 c − c2 a − c2 b) + (a3 b3 + b3 c3 + c3 a3 + 3a2 b2 c2 − a3 b2 c − a3 bc2 − ab3 c2 − ab2 c3 − a2 b3 c − a2 bc3 ) ≥ 0. But this is true from Schur’s Inequality applied for a, b, c and ab, bc, ca. 45. Let a0 =

1 a2 1 and ak+1 = ak + k . Prove that 1 − < an < 1. 2 n n TST Singapore

Solution: 1 1 1 We have and so = − ak+1 ak ak + n  n−1 n−1 X 1 X 1 1 1 − = <1⇒2− < 1 ⇒ an < 1. ak ak+1 n + ak an k=0

k=0

n−1

X 1 1 n = > an n + ak n+1 k=0 and from this inequality and the previous one we conclude immediately that 1 1 − < an < 1. n Now, because the sequence is increasing we also have 2 −

46. [ C˘ alin Popa ] Let a, b, c be positive real numbers, with a, b, c ∈ (0, 1) such that ab + bc + ca = 1. Prove that   a b c 3 1 − a2 1 − b2 1 − c2 + + ≥ + + . 1 − a2 1 − b2 1 − c2 4 a b c

Old and New Inequalities

55

Solution: It is known that in every triangle ABC the following identity holds h π X A A B tan tan = 1 and because tan is bijective on 0, we can set a = tan , b = 2 2 2 2 C B tan , c = tan . The condition a, b, c ∈ (0, 1) tells us that the triangle ABC is 2 2 acute-angled. With this substitution, the inequality becomes X

2 tan

A 2

A 1 − tan 2 2

2

≥3

X 1 − tan

A 2 tan 2

A X X 2 ⇔ tan A ≥ 3

1 ⇔ tan A

⇔ tan A tan B tan C(tan A+tan B+tan C) ≥ 3(tan A tan B+tan B tan C+tan C tan A) ⇔ ⇔ (tan A + tan B + tan C)2 ≥ 3(tan A tan B + tan B tan C + tan C tan A), clearly true because tan A, tan B, tan C > 0. 47. [ Titu Andreescu, Gabriel Dospinescu ] Let x, y, z ≤ 1 and x + y + z = 1. Prove that 1 1 27 1 . + + ≤ 2 2 2 1+x 1+y 1+z 10 Solution: Using the fact that (4 − 3t)(1 − 3t)2 ≥ 0 for any t ≤ 1, we find that 1 27 ≤ (2 − x). 2 1+x 50 Writing two similar expressions for y and z and adding them up, we find the desired inequality. Remark. Tough it may seem too easy, this problem helps us to prove the following difficult inequality X (b + c − a)2 3 ≥ . a2 + (b + c)2 5 In fact, this problem is equivalent to that difficult one. Try to prove this! 48. [ Gabriel Dospinescu ] Prove that if

√

x+

√

y+

√

z = 1, then

(1 − x)2 (1 − y)2 (1 − z)2 ≥ 215 xyz(x + y)(y + z)(z + x) . Solution: √ √ √ We put a = x, b = y and c = z. Then 1 − x = 1 − a2 = (a + b + c)2 − a2 = (b + c)(2a + b + c). Now we have to prove that ((a + b)(b + c)(c + a)(2a + b + c)(a + 2b + c)(a + b + 2c))2 ≥ 215 a2 b2 c2 (a2 + b2 )(b2 + c2 )(c2 + a2 ). But this inequality is true

56

Solutions

(a + b)4 (this being 8 4 equivalent to (a−b) ≥ 0) and (2a+b+c)(a+2b+c)(a+b+2c) ≥ 8(b+c)(c+a)(a+b).

as it follows from the following true inequalities ab(a2 + b2 ) ≤

49. Let x, y, z be positive real numbers such that xyz = x + y + z + 2. Prove that (1) xy + yz + zx ≥ 2(x + y + z); √ √ 3√ √ (2) x + y + z ≤ xyz. 2 Solution: The initial condition xyz = x + y + z + 2 can be rewritten as follows 1 1 1 + + = 1. 1+x 1+y 1+z Now, let 1 1 1 = a, = b, = c. 1+x 1+y 1+z Then 1−a b+c c+a a+b x= = , y= , z= . a a b c (1) We have b+c c+a c+a a+b a+b b+c · + · + · ≥ xy + yz + zx ≥ 2(x + y + z) ⇔ a b b c c a  b+c c+a a+b +· + ≥ 2 ⇔ a3 + b3 + c3 + 3abc ≥ a b c X ≥ ab(a + b) + bc(b + c) + ca(c + a) ⇔ a(a − b)(a − c) ≥ 0, which is exactly Schur’s Inequality. (2) Here we have √ √ 3√ √ x+ y+ z ≤ xyz ⇔ 2 r r r a b b c c a 3 ⇔ · + · + · ≤ . b+c c+a c+a a+b a+b b+c 2 This can be proved by adding the inequality r   a b 1 a b · ≤ + , b+c c+a 2 a+c b+c with the analogous ones. 50. Prove that if x, y, z are real numbers such that x2 + y 2 + z 2 = 2, then x + y + z ≤ xyz + 2. IMO Shortlist, 1987

Old and New Inequalities

57

First solution: If one of x, y, z is negative, let us say x then 2 + xyz − x − y − z = (2 − y − z) − x(1 − yz) ≥ 0, p z2 + y2 because y + z ≤ 2(y 2 + z 2 ) ≤ 2 and zy ≤ ≤ 1. So we may assume that 2 0 < x ≤ y ≤ z. If z ≤ 1 then 2 + xyz − x − y − z = (1 − z)(1 − xy) + (1 − x)(1 − y) ≥ 0. Now, if z > 1 we have z + (x + y) ≤

p

2(z 2 + (x + y)2 ) = 2

p 1 + xy ≤ 2 + xy ≤ 2 + xyz.

This ends the proof. Second solution: Using the Cauchy-Schwarz Inequality, we find that p x + y + z − xyz = x(1 − yz) + y + z ≤ (x2 + (y + z)2 ) · (1 + (1 − yz)2 ). So, it is enough to prove that this last quantity is at most 2, which is equivalent to the inequality (2 + 2yz)(2 − 2yz + (yz)2 ) ≤ 4 ⇔ 2(yz)3 ≤ 2(yz)2 , which is clearly true, because 2 ≥ y 2 + z 2 ≥ 2yz. 51. [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x1 , x2 , . . . , xn ∈ (0, 1) and for any permutation σ of the set {1, 2, . . . , n}, we have the inequality   n X ! xi   n n X X   1 1 i=1   ≥ 1 + · . 1 − xi n  1 − xi · xσ(i)   i=1 i=1

Solution: Using the AM-GM Inequality and the fact that

1 1 1 ≤ + , we can x+y 4x 4y

write the following chain of inequalities n X i=1

1 1 − xi · yi

n X 1 1 = 2 ≤ 2 2 2 1 − xi + 1 − yi2 x + yi i=1 1 − i i=1 2  X n  n X 1 1 1 = + . ≤ 2) 2) 2(1 − x 2(1 − y 1 − x2i i i i=1 i=1

≤

n X

58

Solutions

So, it is enough to prove the inequality   n X xi   n X   1 i=1  · ≥ 1 +  1 − x n i   i=1

⇔

n X i=1

1 xi ≥ · 1 − x2i n

which is Chebyshev Inequality   1 1 1 , ,..., . 1 − x21 1 − x22 1 − x2n

for

n X i=1

1 1 − x2i

n X

! xi

i=1

the

·

! ⇔

n X i=1

systems

52. Let x1 , x2 , . . . , xn be positive real numbers such that

1 1 − x2i

n X √ i=1

xi ≥ (n − 1)

,

(x1 , x2 , . . . , xn )

n X i=1

that

!

and

1 = 1. Prove 1 + xi

n X 1 √ . xi i=1

Vojtech Jarnik First solution: 1 = ai . The inequality becomes Let 1 + xi n r n r n X X X 1 − ai ai 1 p ≥ ≥ (n − 1) ⇔ a 1 − a ai (1 − ai ) i i i=1 i=1 i=1 ! ! n n r n n r X X X X ai 1 ai p ⇔ n ≤ ai . ≥n 1 − ai 1 − ai ai (1 − ai ) i=1 i=1 i=1 i=1 But the last inequality is a consequence of Chebyshev’s Inequality for the n-tuples (a1 , a2 , . . . , an ) and ! 1 1 1 p ,p ,..., p . a1 (1 − a1 ) a2 (1 − a2 ) an (1 − an ) Solution 2 : With the same notations, we have to prove that n r X ai (n − 1) ≤ a + a + . . . + a 1 2 i−1 + ai+1 + . . . + an i=1 ≤

n X i=1

r

a1 + a2 + . . . + ai−1 + ai+1 + . . . + an . ai

Old and New Inequalities

59

But using the Cauchy-Schwarz Inequality and the AM-GM Inequality, we deduce that n r X a1 + a2 + . . . + ai−1 + ai+1 + . . . + an ≥ ai i=1 ≥

= ≥ ≥

√ √ √ √ n √ X a1 + a2 + . . . + ai−1 + ai+1 + . . . + an √ = √ n − 1 · ai i=1   √ n X ai 1 1 1 1 1 √ +√ + ··· + √ ≥ √ + √ + ··· + √ a1 a2 ai−1 ai+1 an n−1 i=1 √ √ n X (n − 1) n − 1 · ai √ √ √ ≥ √ √ a1 + a2 + · · · + ai−1 + ai+1 + · · · + an i=1 r n X ai (n − 1) a + a + · · · + a + ai+1 + · · · + an 1 2 i−1 i=1

and we are done. 53. [ Titu Andreescu ] Let n > 3 and a1 , a2 , . . . , an be real numbers such that a1 + a2 + . . . + an ≥ n and a21 + a22 + . . . + a2n ≥ n2 . Prove that max{a1 , a2 , . . . , an } ≥ 2. USAMO, 1999 Solution: The most natural idea is to suppose that ai < 2 for all i and to substitute n n X X xi = 2 − ai > 0. Then we have (2 − xi ) ≥ n ⇒ xi ≤ n and also i=1 2

n ≤

n X

a2i

=

n X

i=1

i=1 2

(2 − xi ) = 4n − 4

i=1

n X

xi +

n X

i=1 n X

Now, using the fact that xi > 0, we obtain

x2i <

i=1

n X

x2i

i=1

!2 xi

, which combined with

i=1

the above inequality yields n2 < 4n − 4

n X

xi +

i=1

(we have used the fact that

n X

n X

!2 xi

i=1

< 4n + (n − 4)

n X

xi ≤ n). Thus, we have (n − 4)

i=1

which is clearly impossible since n ≥ 4 and

n X i=1

and consequently max{a1 , a2 , . . . , an } ≥ 2.

xi

i=1 n X

! xi − n

> 0,

i=1

xi ≤ n. So, our assumption was wrong

60

Solutions

54. [ Vasile Cˆırtoaje ] If a, b, c, d are positive real numbers, then a−b b−c c−d d−a + + + ≥ 0. b+c c+d d+a a+b

Solution: We have a−b b−c c−d d−a a+c b+d c+a d+b + + + = + + + −4= b+c c+d d+a a+b b+c c+d d+a a+b     1 1 1 1 + + (b + d) + − 4. = (a + c) b+c d+a c+d a+b Since 1 1 4 1 1 4 + ≥ , + ≥ , b+c d+a (b + c) + (d + a) c + d a + b (c + d) + (a + b) we get b−c c−d d−a 4(a + c) 4(b + d) a−b + + + ≥ + − 4 = 0. b+c c+d d+a a+b (b + c) + (d + a) (c + d) + (a + b) Equality holds for a = c and b = d. Conjecture (Vasile Cˆırtoaje) If a, b, c, d, e are positive real numbers, then b−c c−d d−e e−a a−b + + + + ≥ 0. b+c c+d d+e e+a a+b 55. If x and y are positive real numbers, show that xy + y x > 1. France, 1996 Solution: We will prove that ab ≥

a for any a, b ∈ (0, 1). Indeed, from Bernoulli a + b − ab Inequality it follows that a1−b = (1 + a − 1)1−b ≤ 1 + (a − 1)(1 − b) = a + b − ab and thus the conclusion. Now, it x or y is at least 1, we are done. Otherwise, let 0 < x, y < 1. In this case we apply the above observation and find that xy + y x ≥ x y x y + > + = 1. x + y − xy x + y − xy x+y x+y 56. Prove that if a, b, c > 0 have product 1, then (a + b)(b + c)(c + a) ≥ 4(a + b + c − 1). MOSP, 2001

Old and New Inequalities

61

First solution: Using the identity (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) − 1 we reduce the problem to the following one ab + bc + ca +

3 ≥ 4. a+b+c

Now, we can apply the AM-GM Inequality in the following form s (ab + bc + ca)3 3 ab + bc + ca + ≥44 . a+b+c 9(a + b + c) And so it is enough to prove that (ab + bc + ca)3 ≥ 9(a + b + c). But this is easy, because we clearly have ab + bc + ca ≥ 3 and (ab + bc + ca)2 ≥ 3abc(a + b + c) = 3(a + b + c). Second solution: 8 We will use the fact that (a + b)(b + c)(c + a) ≥ (a + b + c)(ab + bc + ca). 9 2 1 So, it is enough to prove that (ab + bc + ca) + ≥ 1. Using the AM-GM 9 a+b+c Inequality, we can write s 1 (ab + bc + ca)2 2 (ab + bc + ca) + ≥33 ≥ 1, 9 a+b+c 81(a + b + c) because (ab + bc + ca)2 ≥ 3abc(a + b + c) = 3(a + b + c).

57. Prove that for any a, b, c > 0, (a2 + b2 + c2 )(a + b − c)(b + c − a)(c + a − b) ≤ abc(ab + bc + ca).

Solution: Clearly, if one of the factors in the left-hand side is negative, we are done. So, we may assume that a, b, c are the side lenghts of a triangle ABC. With the usual notations in a triangle, the inequality becomes (a2 +b2 +c2 )·

16K 2 ≤ abc(ab+bc+ca) ⇔ (a+b+c)(ab+bc+ca)R2 ≥ abc(a2 +b2 +c2 ). a+b+c

But this follows from the fact that (a + b + c)(ab + bc + ca) ≥ 9abc and 0 ≤ OH 2 = 9R2 − a2 − b2 − c2 .

62

Solutions

58. [ D.P.Mavlo ] Let a, b, c > 0. Prove that 3+a+b+c+

1 1 1 a b c (a + 1)(b + 1)(c + 1) + + + + + ≥3 . a b c b c a 1 + abc Kvant, 1988

Solution: The inequality is equivalent to the following one X X X X1 Xa ab + a a+ + ≥3 a b abc + 1 or X Xa X1 X X  a2 c + + ≥2 a+ ab . a b But this follows from the inequalities abc

X

a+

a2 bc +

c a b ≥ 2ab, b2 ca + ≥ 2bc, c2 ab + ≥ 2ca c a b

and a2 c +

1 1 1 ≥ 2a, b2 a + ≥ 2b, c2 b + ≥ 2c. c a b

59. [ Gabriel Dospinescu ] Prove that for any positive real numbers x1 , x2 , . . . , xn with product 1 we have the inequality !n n n n Y X X 1 n n . n · (xi + 1) ≥ xi + x i=1 i=1 i=1 i

Solution: Using the AM-GM Inequality, we deduce that xnn−1 xn1 xn2 1 + + ··· + + ≥ n n 1 + x1 1 + x2 1 + xnn−1 1 + xnn

n v uY n u n (1 + xni ) xn · t i=1

and

1 1 1 xnn n · xn + + ··· + + ≥ v n n n n u 1 + x1 1 + x2 1 + xn−1 1 + xn n uY n t (1 + xni ) i=1

Thus, we have v u n uY 1 n t (1 + xni ) ≥ xn + . x n i=1

Old and New Inequalities

63

Of course, this is true for any other variable, so we can add all these inequalities to obtain that v u n n n X X uY 1 n nt (1 + xni ) ≥ xi + x i=1 i=1 i=1 i which is the desired inequality. 60. Let a, b, c, d > 0 such that a + b + c = 1. Prove that   1 1 d 3 3 3 a + b + c + abcd ≥ min , + . 4 9 27 Kvant, 1993 Solution: 1 , Suppose the inequality is false. Then we have, taking into account that abc ≤ 27   1 1 1 the inequality d − abc > a3 + b3 + c3 − . We may assume that abc < . Now, 27 9 27 1 we will reach a contradiction proving that a3 + b3 + c3 + abcd ≥ . It is sufficient to 4 prove that 1 a3 + b3 + c3 − 9 abc + a3 + b3 + c3 ≥ 1 . 1 4 − abc 27 X But this inequality is equivalent to 4 a3 + 15abc ≥ 1. We use now the identity X X X 1 + 9abc , a3 = 3abc+1−3 ab and reduce the problem to proving that ab ≤ 4 which is Schur’s Inequality. 61. Prove that for any real numbers a, b, c we have the inequality X

(1 + a2 )2 (1 + b2 )2 (a − c)2 (b − c)2 ≥ (1 + a2 )(1 + b2 )(1 + c2 )(a − b)2 (b − c)2 (c − a)2 . AMM Solution: The inequality can be also written as

X (1 + a2 )(1 + b2 )

(1 + c2 )(a − b)2 assume that a, b, c are distinct). Now, adding the inequalities

≥ 1 (of course, we may

(1 + a2 )(1 + b2 ) (1 + b2 )(1 + c2 ) 1 + b2 + ≥ 2 (1 + c2 )(a − b)2 (1 + a2 )(b − c)2 |a − b||c − b| (which can be found using the AM-GM Inequality) we deduce that X (1 + a2 )(1 + b2 ) (1 +

c2 )(a

−

b)2

≥

X

1 + b2 |(b − a)(b − c)|

64

Solutions

and so it is enough to prove that the last quantity is at least 1. But it follows from X X 1 + b2 1 + b2 =1 ≥ |(b − a)(b − c)| (b − a)(b − c) and the problem is solved. 62. [ Titu Andreescu, Mircea Lascu ] Let α, x, y, z be positive real numbers such that xyz = 1 and α ≥ 1. Prove that xα yα zα 3 + + ≥ . y+z z+x x+y 2

First solution: We may of course assume that x ≥ y ≥ z. Then we have y z x ≥ ≥ y+z z+x x+y and xα−1 ≥ y α−1 ≥ z α−1 . Using Chebyshev’s Inequality we infer that   X xα 1 X α−1  X x ≥ · x · . y+z 3 y+z X Now, all we have to do is to observe that this follows from the inequalities xα−1 ≥ 3 X x 3 ≥ (from the AM-GM Inequality) and y+z 2 Second solution: According to the Cauchy-Schwarz Inequality, we have:    α 2 1+α 1+α 1+α yα zα x + + ≥ x 2 +y 2 +z 2 . [(x(y +z)+y(z +x)+z(x+y)]] y+z z+x x+y Thus it remains to show that  1+α 2 1+α 1+α x 2 +y 2 +z 2 ≥ 3(xy + yz + zx). Since (x + y + z)2 ≥ 3(xy + yz + zx), it is enough to prove that x

1+α 2

+y

1+α 2

+z

1+α 2

≥ x + y + z.

From Bernoulli’s Inequality, we get x

1+α 2

= [1 + (x − 1)]

1+α 2

≥1+

1+α 1−α 1+α (x − 1) = + x 2 2 2

and, similarly, y

1+α 2

≥

1+α 1−α 1+α 1−α 1+α + y, z 2 ≥ + z. 2 2 2 2

Thus x

1+α 2

+y

1+α 2

+z

1+α 2

− (x + y + z) ≥

3(1 − α) 1 + α + (x + y + z) − (x + y + z) = 2 2

Old and New Inequalities

65

α−1 α−1 √ (x + y + z − 3) ≥ (3 3 xyz − 3) = 0. 2 2 Equality holds for x = y = z = 1. Remark : 1 1 1 Using the substitution β = α + 1 (β ≥ 2) and x = , y = , z = (abc = 1) the a b c inequality becomes as follows 1 1 1 3 + β + γ ≥ . β a (b + c) b (c + a) c (a + b) 2 For β = 3, we obtain one of the problems of IMO 1995 (proposed by Russia).

y12

63. Prove that for any real numbers x1 , . . . , xn , y1 , . . . , yn such that x21 +· · ·+x2n = + · · · + yn2 = 1, ! n X (x1 y2 − x2 y1 )2 ≤ 2 1 − xk yk . k=1

Korea, 2001 Solution: We clearly have the inequality 2

(x1 y2 − x2 y1 ) ≤

X

2

(xi yj − xj yi )

=

n X

! x2i

i=1

1≤i<j≤n

=

1−

n X

! yi2

−

i=1 n X

!2 xi yi

=

i=1

! xi yi

n X

1+

i=1

n X

! xi yi

.

i=1

n X Because we also have xi yi ≤ 1, we find immediately that i=1 ! ! ! n n n X X X xi yi xi yi ≤ 2 1 − 1− xi yi 1+ i=1

i=1

i=1

and the problem is solved. 64. [ Laurent¸iu Panaitopol ] Let a1 , a2 , . . . , an be pairwise distinct positive integers. Prove that 2n + 1 a21 + a22 + · · · + a2n ≥ (a1 + a2 + · · · + an ). 3 TST Romania

66

Solutions

Solution: Without loss of generality, we may assume that a1 < a2 < · · · < an and hence ai ≥ i for all i. Thus, we may take bi = ai − i ≥ 0 and the inequality becomes n X i=1

b2i + 2

n X

ibi +

i=1

n 2n + 1 X n(n + 1)(2n + 1) n(n + 1)(2n + 1) ≥ · . bi + 6 3 6 i=1

Now, using the fact that ai+1 > ai we infer that b1 ≤ b2 ≤ · · · ≤ bn and from Chebyshev’s Inequality we deduce that 2

n X

ibi ≥ (n + 1)

i=1

n X i=1

n

bi ≥

2n + 1 X bi 3 i=1

and the conclusion is immediate. Also, from the above relations we can see immediately that we have equality if and only if a1 , a2 , . . . , an is a permutation of the numbers 1, 2, . . . , n. 65. [ C˘ alin Popa ] Let a, b, c be positive real numbers such that a + b + c = 1. Prove that √ √ √ √ b c c a a b 3 3 √ √ √ . + √ + √ ≥ √ 4 a( 3c + ab) b( 3a + bc) c( 3b + ca)

Solution: Rewrite the inequality in the form r

bc √ 3 3 r a ≥ . 4 3ca +a b r r r bc ca ab ,y = ,z = the condition a + b + c = 1 With the substitution x = a b c becomes xy + yz + zx = 1 and the inequality turns into √ X 3 3 x √ ≥ . 4 3y + yz X

But, by applying the Cauchy-Schwarz Inequality we obtain X 2 X √ x 2 X 3 xy x 3 3 √ ≥√ ≥√ = , 1 4 3xy + xyz 3 + 3xyz 3+ √ 3 where we used the inequalities X 2 X  1 x ≥3 xy and xyz ≤ √ . 3 3

Old and New Inequalities

67

66. [ Titu Andreescu, Gabriel Dospinescu ] Let a, b, c, d be real numbers such that (1 + a2 )(1 + b2 )(1 + c2 )(1 + d2 ) = 16. Prove that −3 ≤ ab + bc + cd + da + ac + bd − abcd ≤ 5.

Solution: Y Y Let us write the condition in the form 16 = (i + a) · (a − i). Using symmetric sums, we can write this as follows    X X X X X X 16 = 1 − i a− ab + i abc + abcd 1 + i a− ab − i abc + abcd . P P P So, we have the identity 16 = (1 − ab + abcd)2 + ( a − abc)2 . This means that P |1 − ab + abcd| ≤ 4 and from here the conclusion follows. 67. Prove that (a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca) for any positive real numbers a, b, c. APMO, 2004 First solution: We will prove even more: (a2 + 2)(b2 + 2)(c2 + 2) ≥ 3(a + b + c)2 . Because (a + b + c)2 ≤ (|a| + |b| + |c|)2 , we may assume that a, b, c are nonnegative. We will use the fact that if x and y have the same sign then (1 + x)(1 + y) ≥ 1 + x + y. So, we write the inequality in the form  Y  a2 − 1 (a + b + c)2 +1 ≥ 3 9 and we have three cases

X 2  a X a2 − 1 i) If a, b, c are at least 1, then +1 ≥1+ ≥ . 3 3 9 ii) If two of the three numbers are at least 1, let them be a and b, then we have Y  a2 − 1

Y  a2 − 1 3

 +1

 ≥

a2 − 1 b2 − 1 1+ + 3 3



c2 + 2 3



(a2 + b2 + 1) (12 + 12 + c2 ) (a + b + c)2 ≥ 9 9 9 by the Cauchy-Schwarz Inequality. iii) If all three numbers are at most 1, then by Bernoulli Inequality we have X 2  a Y  a2 − 1 X a2 − 1 +1 ≥1+ ≥ 3 3 9 =

and the proof is complete.

68

Solutions

Second solution: Expanding everything, we reduce the problem to proving that X X X (abc)2 + 2 a2 b2 + 4 a2 + 8 ≥ 9 ab. X X X X Because 3 a2 ≥ 3 ab and 2 a2 b2 + 6 ≥ 4 ab, we are left with the X X 2 2 inequality (abc) + a +2 ≥ 2 ab. Of course, we can assume that a, b, c are 2 non-negative and we can write a = x , b = y 2 , c = z 2 . In this case X X 2 ab − a2 = (x + y + z)(x + y − z)(y + z − x)(z + x − y). It is clear that if x, y, z are not side lengths of a triangle, then the inequality is trivial. Otherwise, we can take x = u + v, y = v + w, z = w + u and reduce the inequality to 4

((u + v)(v + w)(w + u)) + 2 ≥ 16(u + v + w)uvw. p 4 We have ((u + v)(v + w)(w + u)) + 1 + 1 ≥ 3 3 (u + v)4 (v + w)4 (u + w)4 and it remains to prove that the last quantity is at least 16(u + v + w)uvw. This comes down to 163 (u + v)4 (v + w)4 (w + u)4 ≥ 3 (uvw)3 (u + v + w)3 . 3 But this follows from the known inequalities 8 (u + v)(v + w)(w + u) ≥ (u + v + w)(uv + vw + wu), 9 8 √ (uv + vw + wu)4 ≥ 34 (uvw) 3 , u + v + w ≥ 3 3 uvw. Third solution: In the same manner as in the Second solution, we reduce the problem to proving that X X (abc)2 + 2 ≥ 2 ab − a2 . Now, using Schur’s Inequality, we infer that X X 2 ab − a2 ≤

9abc a+b+c and as an immediate consequence of the AM-GM Inequality we have p 9abc ≤ 3 3 (abc)2 . a+b+c This shows that as soon as we prove that p (abc)2 + 2 ≥ 3 3 (abc)2 , the problem is solved. But the last assertion follows from the AM-GM Inequality.

Old and New Inequalities

69

68. [ Vasile Cˆırtoaje ] Prove that if 0 < x ≤ y ≤ z and x + y + z = xyz + 2, then a) (1 − xy)(1 − yz)(1 − xz) ≥ 0; 32 b) x2 y ≤ 1, x3 y 2 ≤ . 27 Solution: a) We have (1 − xy)(1 − yz) = 1 − xy − yz + xy 2 z = 1 − xy − yz + y(x + y + z − 2) = (y − 1)2 ≥ 0 and similarly (1 − yz)(1 − zx) = (1 − z)2 ≥ 0,

(1 − zx)(1 − xy) = (1 − x)2 ≥ 0.

So the expressions 1 − xy, 1 − yz and 1 − zx have the same sign. b) We rewrite the relation x+y +z = xyz +2 as (1−x)(1−y)+(1−z)(1−xy) = 0. If x > 1 then z ≥ y ≥ x > 1 and so (1 − x)(1 − y) + (1 − z)(1 − xy) > 0, impossible. So we have x ≤ 1. Next we distinguish two cases 1) xy ≤ 1; 2) xy > 1. 32 . 1) xy ≤ 1. We have x2 y ≤ x ≤ 1 and x3 y 2 ≤ x ≤ 1 < 27 √ 2) xy > 1. From y ≥ xy we get y > 1. Next we rewrite the relation x + y + z = xyz + 2 as x + y − 2 = (xy − 1)z. Because z ≥ y gives x + y − 2 ≥ (xy − 1)y, (y − 1)(2 − x − xy) ≥ 0 so 2 ≥ x(1 + y). Using r the AM-GM Inequality, we have y y y y √ √ 1 + y ≥ 2 y and 1 + y = 1 + + ≥ 3 3 1 · · . Thus we have 2 ≥ 2x y and 2 2 2 2 r 2 32 3 y 2 ≥ 3x , which means that x2 y ≤ 1 and x3 y 2 ≤ . 4 27 32 The equality x2 y = 1 takes place when x = y = 1 and the equality x3 y 2 = 27 2 takes place when x = , y = z = 2. 3 69. [ Titu Andreescu ] Let a, b, c be positive real numbers such that a+b+c ≥ abc. Prove that at least two of the inequalities 2 3 6 2 3 6 2 3 6 + + ≥ 6, + + ≥ 6, + + ≥ 6, a b c b c a c a b are true. TST 2001, USA Solution: 1 1 1 The most natural idea is to male the substitution = x, = y, = z. Thus, we a b c have x, y, z > 0 and xy + yz + zx ≥ 1 and we have to prove that at least two of the inequalities 2x+3y+6z ≥ 6, 2y+3z +6x ≥ 6, 2z +3x+6y ≥ 6 are true. Suppose this is not the case. Then we may assume that 2x+3y +6z < 6 and 2z +3x+6y < 6. Adding, 5 − 5yz 1 − yz we find that 5x+9y +8z < 12. But we have x ≥ . Thus, 12 > +9y +8z y+z y+z

70

Solutions

which is the same as 12(y + z) > 5 + 9y 2 + 8z 2 + 12yz ⇔ (2z − 1)2 + (3y + 2z − 2)2 < 0, which is clearly impossible. Thus, the conclusion follows. 70. [ Gabriel Dospinescu, Marian Tetiva ] Let x, y, z > 0 such that x + y + z = xyz. Prove that

√ (x − 1) (y − 1) (z − 1) ≤ 6 3 − 10.

First solution: Because of x < xyz ⇒ yz > 1 (and the similar relations xz > 1, xy > 1) at most one of the three numbers can be less than 1. In any of these cases (x ≤ 1, y ≥ 1, z ≥ 1 or the similar ones) the inequality to prove is clear. The only case we still have to analyse is that when x ≥ 1, y ≥ 1 and z ≥ 1. In this situation denote x − 1 = a, y − 1 = b, z − 1 = c. Then a, b, c are nonnegative real numbers and, because x = a + 1, y = b + 1, z = c + 1, they satisfy a + 1 + b + 1 + c + 1 = (a + 1) (b + 1) (c + 1) , which means Now let x =

√ 3

abc + ab + ac + bc = 2. abc; we have √ 3 ab + ac + bc ≥ 3 abacbc = 3x2 ,

that’s why we get
x3 + 3x2 ≤ 2 ⇔ (x + 1) x2 + 2x − 2 ≤ 0 ⇔ √
√
⇔ (x + 1) x + 1 + 3 x + 1 − 3 ≤ 0. For x ≥ 0, this yields

√ 3

abc = x ≤

or, equivalently abc ≤ which is exactly The proof is complete.

√

√

3 − 1,

3 3−1 ,

√ (x − 1) (y − 1) (z − 1) ≤ 6 3 − 10.

Old and New Inequalities

71

Second solution: Like in the first solution (and due to the symmetry) we may suppose that x ≥ 1, y ≥ 1; we can even assume that x > 1, y > 1 (for x = 1 the inequality is plain). Then we get xy > 1 and from the given condition we have x+y z= . xy − 1 The relation to prove is √ (x − 1) (y − 1) (z − 1) ≤ 6 3 − 10 ⇔ √ ⇔ 2xyz − (xy + xz + yz) ≤ 6 3 − 9, or, with this expression of z, √ x+y x+y − xy − (x + y) ≤6 3−9⇔ 2xy xy − 1 √ √ xy −
1 2 ⇔ (xy − x − y) + 6 3 − 10 xy ≤ 6 3 − 9, after some calculations. Now, we put x = a + 1, y = b + 1 and transform this into  √  a2 b2 + 6 3 − 10 (a + b + ab) − 2ab ≥ 0. But

√ a + b ≥ 2 ab

√ and 6 3 − 10 > 0, so it suffices to show that  √   √ a2 b2 + 6 3 − 10 2 ab + ab − 2ab ≥ 0. √ The substitution t = ab ≥ 0 reduces this inequality to  √   √  t4 + 6 3 − 12 t2 + 2 6 3 − 10 t ≥ 0, or

 √   √  t3 + 6 3 − 12 t + 2 6 3 − 10 ≥ 0.

The derivative of the function  √   √  f (t) = t3 + 6 3 − 12 t + 2 6 3 − 10 , t ≥ 0 is

 √ 2  0 f (t) = 3 t2 − 3−1 √ and has only one positive zero. It is 3 − 1 and it’s easy to see that this is a minimum point for f in the interval [0, ∞). Consequently √  f (t) ≥ f 3 − 1 = 0, and we are done. A final observation: in fact we have  2   √ √ f (t) = t − 3 + 1 t+2 3−2 ,

72

Solutions

which shows that f (t) ≥ 0 for t ≥ 0. 71. [ Marian Tetiva ] Prove that for any positive real numbers a, b, c, 3 a − b3 b3 − c3 c3 − a3 (a − b)2 + (b − c)2 + (c − a)2 ≤ + + . a+b b+c c+a 4 Moldova TST, 2004 First solution: First of all, we observe that right hand side can be transformed into     X a3 − b3 1 1 1 1 = (a3 − b3 ) − + (b3 − c3 ) − = a+b a+b a+c b+c a+c X  (a − b)(c − b)(a − c) ab = (a + b)(a + c)(b + c) and so we have to prove the inequality 1 X 2 X  |(a − b)(b − c)(c − a)|(ab + bc + ca) ≤ a − ab . (a + b)(b + c)(c + a) 2 8 It is also easy to prove that (a + b)(b + c)(c + a) ≥ (a + b + c)(ab + bc + ca) and 9 so we are left with  Y 2 X X · a· (a − b)2 ≥ (a − b) . 9 Using the AM-GM Inequality, we reduce this inequality to the following one 8 X 3 Y a ≥ (a − b) . 27 This one is easy. Just observe that we can assume that a ≥ b ≥ c and in this case it becomes 8 (a − b)(a − c)(b − c) ≤ (a + b + c)3 27 and it follows from the AM-GM Inequality. Second solution (by Marian Tetiva): It is easy to see that the inequality is not only cyclic, but symmetric. That is why we may assume that a ≥ b ≥ c > 0. The idea is to use the inequality y x2 + xy + y 2 x ≥ ≥y+ , 2 x+y 2 which is true if x ≥ y > 0. The proof of this inequality is easy and we won’t insist. Now, because a ≥ b ≥ c > 0, we have the three inequalities x+

a+

b a2 + ab + b2 a c b2 + bc + c2 b ≥ ≥ b + ,b + ≥ ≥c+ 2 a+b 2 2 b+c 2

and of course a+

c a2 + ac + c2 a ≥ ≥c+ . 2 a+c 2

Old and New Inequalities

73

That is why we can write X a3 − b3 a2 + ab + b2 b2 + bc + c2 a2 + ac + c2 = (a − b) + (b − c) − (a − c) ≥ a+b a+b b+c a+c     a b c ≥ (a − b) b + + (b − c) c + = − (a − c) a + 2 2 2 X (a − b)2 = − . 4 In the same manner we can prove that P X a3 − b3 (a − b)2 ≤ a+b 4 and the conclusion follows. 72. [ Titu Andreescu ] Let a, b, c be positive real numbers. Prove that (a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a + b + c)3 . USAMO, 2004 Solution: We start with the inequality a5 − a2 + 3 ≥ a3 + 2 ⇔ (a2 − 1)(a3 − 1) ≥ 0. Thus, it remains to show that Y
X 3 a . a3 + 2 ≥ Using the AM-GM Inequality, one has 1 1 3a a3 + + ≥ qY . + 2 b3 + 2 c3 + 2 3 (a3 + 2)

a3

We write two similar inequalities and then add up all these relations. We will find that X 3 Y (a3 + 2) ≥ a , which is what we wanted. 73. [ Gabriel Dospinescu ] Let n > 2 and x1 , x2 , . . . , xn > 0 such that ! ! n n X X 1 = n2 + 1. xk xk k=1

Prove that

n X k=1

! x2k

·

k=1

n X 1 x2k

k=1

! > n2 + 4 +

2 . n(n − 1)

74

Solutions

Solution: In this problem, a combination between identities and the Cauchy-Schwarz Inequality is the way to proceed. So, let us start with the expression 2 X  xi xj + −2 . xj xi 1≤i<j≤n

We can immediately see that 2 X  xi xj + −2 = xj xi

! x2j xi x2i xj + 2 −4· −4· +6 = x2j xi xj xi 1≤i<j≤n ! n ! ! ! n n n X X 1 X X 1 2 = xi −4 xi + 3n2 . 2 x x i i=1 i=1 i i=1 i=1

1≤i<j≤n

X

Thus we could find from the inequality 2 X  xi xj + −2 ≥0 xj xi 1≤i<j≤n

that n X

! x2i

i=1

n X 1 x2 i=1 i

! ≥ n2 + 4.

Unfortunately, this is not enough. So, let us try to minimize 2 X  xi xj + −2 . xj xi 1≤i<j≤n

This could be done using the Cauchy-Schwarz Inequality:    2 X  xi xj  + −2    x xi 2 j X 1≤i<j≤n xj xi   + −2 ≥ . n xj xi 1≤i<j≤n 2   X xj xi + − 2 = 1, we deduce that Because xj xi i≤i<j≤n

n X

! x2i

i=1

n X 1 2 x i=1

! ≥ n2 + 4 +

2 , n(n − 1)

which is what we wanted to prove. Of course, we should prove that we cannot have equality. But equality would imply that x1 = x2 = · · · = xn , which contradicts the assumption ! n ! n X X 1 xk = n2 + 1. xk k=1

k=1

Old and New Inequalities

75

74. [ Gabriel Dospinescu, Mircea Lascu, Marian Tetiva ] Prove that for any positive real numbers a, b, c, a2 + b2 + c2 + 2abc + 3 ≥ (1 + a)(1 + b)(1 + c).

First solution: Let f (a, b, c) = a2 + b2 + c2 + abc + 2 − a − b − c − ab − bc − ca. We have to prove 1 1 1 that all values of f are nonnegative. If a, b, c > 3, then we clearly have + + < 1, a b c which means that f (a, b, c) > a2 + b2 + c2 + 2 − a − b − c > 0. So, we may assume b+c that a ≤ 3 and let m = . Easy computations show that f (a, b, c) − f (a, m, m) = 2 2 (3 − a)(b − c) ≥ 0 and so it remains to prove that f (a, m, m) ≥ 0, which is the same 4 as (a + 1)m2 − 2(a + 1)m + a2 − a + 2 ≥ 0. This is clearly true, because the discriminant of the quadratic equation is −4(a + 1)(a − 1)2 ≤ 0. Second solution: Recall Turkevici’s Inequality x4 + y 4 + z 4 + t4 + 2xyzt ≥ x2 y 2 + y 2 z 2 + z 2 t2 + t2 x2 + x2 z 2 + y 2 t2 for all positive real numbers x, y, z, t. Taking t = 1, a = x2 , b = y 2 , x = z 2 and using √ the fact that 2 abc ≤ abc + 1, we find the desired inequality. 75. [ Titu Andreescu, Zuming Feng ] Let a, b, c be positive real numbers. Prove that (2a + b + c)2 (2b + a + c)2 (2c + a + b)2 + + ≤ 8. 2a2 + (b + c)2 2b2 + (a + c)2 2c2 + (a + b)2 USAMO, 2003 First solution: Because the inequality is homogeneous, we can assume that a + b + c = 3. Then (2a + b + c)2 1 4a + 3  a2 + 6a + 9 = = 1 + 2 · ≤ 2a2 + (b + c)2 3a2 − 6a + 9 3 2 + (a − 1)2 1 4a + 3  4a + 4 ≤ 1+2· = . 3 2 3 Thus X (2a + b + c)2 1X (4a + 4) = 8. ≤ 2a2 + (b + c)2 3

76

Solutions

Second solution: b+c c+a a+b Denote x = , y= ,z = . We have to prove that a b c X (x + 2)2 X 2x + 1 X (x − 1)2 5 1 ≤ 8 ⇔ ≤ ⇔ ≥ . x2 + 2 x2 + 2 2 x2 + 2 2 But, from the Cauchy-Schwarz Inequality, we have X (x − 1)2 (x + y + z − 3)2 ≥ . x2 + 2 x2 + y 2 + z 2 + 6 It remains to prove that 2(x2 + y 2 + z 2 + 2xy + 2yz + 2zx − 6x − 6y − 6z + 9) ≥ x2 + y 2 + z 2 + 6 ⇔ ⇔ x2 + y 2 + z 2 + 4(xy + yz + zx) − 12(x + y + z) + 12 ≥ 0. p Now xy + yz + zx ≥ 3 3 x2 y 2 z 2 ≥ 12 (because xyz ≥ 8), so we still have to prove that (x + y + z)2 + 24 − 12(x + y + z) + 12 ≥ 0, which is equivalent to (x + y + z − 6)2 ≥ 0, clearly true. 76. Prove that for any positive real numbers x, y and any positive integers m, n, (n−1)(m−1)(xm+n +y m+n )+(m+n−1)(xm y n +xn y m ) ≥ mn(xm+n−1 y+y m+n−1 x). Austrian-Polish Competition, 1995 Solution: We transform the inequality as follows: mn(x − y)(xm+n−1 − y m+n−1 ) ≥ (m + n − 1)(xm − y m )(xn − y n ) ⇔ xm − y m xn − y n xm+n−1 − y m+n−1 ≥ · (m + n − 1)(x − y) m(x − y) n(x − y) (we have assumed that x > y). The last relation can also be written Z x Z x Z x m+n−2 m−1 (x − y) t dt ≥ t dt · tn−1 dt ⇔

y

y

y

and this follows from Chebyshev’s Inequality for integrals. 77. Let a, b, c, d, e be positive real numbers such that abcde = 1. Prove that b + bcd c + cde d + dea e + eab 10 a + abc + + + + ≥ . 1 + ab + abcd 1 + bc + bcde 1 + cd + cdea 1 + de + deab 1 + ea + eabc 3 Crux Mathematicorum Solution: We consider the standard substitution y z t u x a = ,b = ,c = ,d = ,e = y z t u x

Old and New Inequalities

77

with x, y, z, t, u > 0. It is clear that 1 1 + a + abc y t . = 1 1 1 1 + ab + abcd + + x z u 1 1 1 1 1 Writing the other relations as well, and denoting = a1 , = a2 , = a3 , = a4 , = x y z t u a5 , we have to prove that if ai > 0, then X a2 + a4 1 ≥ . a1 + a3 + a5 3 Using the Cauchy-Schwarz Inequality, we minor the left-hand side with 4S 2 , 2S 2 − (a2 + a4 )2 − (a1 + a4 )2 − (a3 + a5 )2 − (a2 + a5 )2 − (a1 + a3 )2 where S =

5 X

ai . By applying the Cauchy-Schwarz Inequality again for the

i=1

denominator of the fraction, we obtain the conclusion.  π 78. [ Titu Andreescu ] Prove that for any a, b, c, ∈ 0, the following inequality 2 holds sin a · sin(a − b) · sin(a − c) sin b · sin(b − c) · sin(b − a) sin c · sin(c − a) · sin(c − b) + + ≥ 0. sin(b + c) sin(c + a) sin(a + b) TST 2003, USA Solution: Let x = sin a, y = sin b, z = sin c. Then we have x, y, z > 0. It is easy to see that the following relations are true: sin a · sin(a − b) · sin(a − c) · sin(a + b) · sin(a + c) = x(x2 − y 2 )(x2 − z 2 ) Using similar relations for the other terms, we have to prove that: X x(x2 − y 2 )(x2 − y 2 ) ≥ 0. X√ √ √ √ With the substitution x = u, y = v, z = w the inequality becomes u(u − v)(u − w) ≥ 0. But this follows from Schur’s Inequality. 79. Prove that if a, b, c are positive real numbers then, p p p p a4 + b4 + c4 + a2 b2 + b2 c2 + c2 a2 ≥ a3 b + b3 c + c3 a + ab3 + bc3 + ca3 . KMO Summer Program Test, 2001 Solution: It is clear that it suffices to prove the following inequalities X X X X a4 + a2 b2 ≥ a3 b + ab3

78

Solutions

and X

a4

 X

 X  X  a2 b2 ≥ a3 b ab3 .

The first one follows from Schur’s Inequality X X X X a4 + abc a≥ a3 b + ab3 and the fact that X

a2 b2 ≥ abc

X

a.

The second one is a simple consequence of the Cauchy-Schwarz Inequality: (a3 b + b3 c + c3 a)2 ≤ (a2 b2 + b2 c2 + c2 a2 )(a4 + b4 + c4 ) (ab3 + bc3 + ca3 )2 ≤ (a2 b2 + b2 c2 + c2 a2 )(a4 + b4 + c4 ). 80. [ Gabriel Dospinescu, Mircea Lascu ] For a given n > 2 find the smallest constant kn with the property: if a1 , . . . , an > 0 have product 1, then a2 a3 an a1 a1 a2 + 2 + ··· + 2 ≤ kn . 2 2 2 (a1 + a2 )(a2 + a1 ) (a2 + a3 )(a3 + a2 ) (an + a1 )(a21 + an )

Solution: Let us take first a1 = a2 = · · · = an−1 = x, an = kn ≥

(xn+1

1 . We infer that xn−1

2x2n−1 n−2 n−2 + > + 1)(x2n−1 + 1) (1 + x)2 (1 + x)2

for all x > 0. Clearly, this implies kn ≥ n − 2. Let us prove that n − 2 is a good constant and the problem will be solved. First, we will prove that (x2 + y)(y 2 + x) ≥ xy(1 + x)(1 + y). Indeed, this is the same as (x + y)(x − y)2 ≥ 0. So, it suffices to prove that 1 1 1 + + ··· + ≤ n − 2. (1 + a1 )(1 + a2 ) (1 + a2 )(1 + a3 ) (1 + an )(1 + a1 ) x1 xn Now, we take a1 = , . . . , an = and the above inequality becomes x2 x1  n  X xk+1 xk+2 1− ≥ 2, (xk + xk+1 )(xk+1 + xk+2 ) k=1

which can be also written in the following from n X k=1

x2k+1 xk + ≥ 2. xk + xk+1 (xk + xk+1 )(xk+1 + xk+2 )

Clearly, n X k=1

n

X xk xk ≥ = 1. xk + xk+1 x1 + x2 + · · · + xn k=1

Old and New Inequalities

79

So, we have to prove the inequality n X k=1

x2k+1 ≥ 1. (xk + xk+1 )(xk+1 + xk+2 )

Using the Cauchy-Schwarz Inequality, we infer that n X n X

x2k+1 ≥ n X (xk + xk+1 )(xk+1 + xk+2 )

k=1

!2 xk

k+1

(xk + xk+1 )(xk+1 + xk+2 )

k=1

and so it suffices to prove that !2 n n n n X X X X xk ≥ x2k + 2 xk xk+1 + xk xk+2 . k=1

k=1

k=1

k=1

But this one is equivalent to X

2

xi xj ≥ 2

1≤i<j≤n

n X

xk xk+1 +

k=1

n X

xk xk+2

k=1

and it is clear. Thus, kn = n − 2. 81. [ Vasile Cˆırtoaje ] For any real numbers a, b, c, x, y, z prove that the inequality holds p 2 ax + by + cz + (a2 + b2 + c2 )(x2 + y 2 + z 2 ) ≥ (a + b + c)(x + y + z). 3 Kvant, 1989 Solution:

r

Let us denote t =

x2 + y 2 + z 2 . Using the substitutions x = tp, y = tq and a2 + b2 + c2

z = tr, which imply a2 + b2 + c2 = p2 + q 2 + r2 . The given inequality becomes 2 (a + b + c)(p + q + r), 3 4 (a + p)2 + (b + q)2 + (c + r)2 ≥ (a + b + c)(p + q + r). 3 ap + bq + cr + a2 + b2 + c2 ≥

Since 4(a + b + c)(p + q + r) ≤ [(a + b + c) + (p + q + r)]2 , it suffices to prove that (a + p)2 + (b + q)2 + (c + r)2 ≥ This inequality is clearly true.

1 [(a + p) + (b + q) + (c + r)]2 . 3

80

Solutions

82. [ Vasile Cˆırtoaje ] Prove that the sides a, b, c of a triangle satisfy the inequality     a b c b c a 3 + + −1 ≥2 + + . b c a a b c First solution: a+c . After We may assume that c is the smallest among a, b, c. Then let x = b − 2 some computations, the inequality becomes  a (a−c)2 ≥ 0 ⇔ (3a−2c)(2b−a−c)2 +(4b+2c−3a)(a−c)2 ≥ 0 (3a−2c)x2 + x + c − 4 which follows immediately from 3a ≥ 2c, 4b + 2c − 3a = 3(b + c − a) + b − c > 0. Second solution: Make the classical substitution a = y + z, b = z + x, c = x + y and clear denominators. The problem reduces to proving that x3 + y 3 + z 3 + 2(x2 y + y 2 z + z 2 x) ≥ 3(xy 2 + yz 2 + zx2 ). We can of course assume that x is the smallest among x, y, z. Then we can write y = x + m, z = x + n with nonnegatives m and n. A short computation shows that the inequality reduces to 2x(m2 − mn + n2 ) + m3 + n3 + 2m2 n − 3n2 m ≥ 0. All we need to prove is that m3 + n3 + 2m2 n ≥ 3n2 m ⇔ (n − m)3 − (n − m)m2 + m3 ≥ 0 and this follows immediately from the inequality t3 + 1 ≥ 3, true for t ≥ −1. 83. [ Walther Janous ] Let n > 2 and let x1 , x2 , . . . , xn > 0 add up to 1. Prove that  Y  n  n  Y 1 n − xi 1+ ≥ . xi 1 − xi i=1 i=1 Crux Mathematicorum First solution: The most natural idea is to use the fact that n−1 n − xi =1+ . 1 − xi x1 + x2 + · · · + xi−1 + xi+1 + · · · + xn Thus, we have  n  Y n − xi i=1

1 − xi

≤

n  Y i=1

1 1 + n−1 √ x1 x2 . . . xi−1 xi+1 . . . xn

and we have to prove the inequality  Y n  n  Y 1 1+ ≥ 1+ xi i=1 i=1

√ n−1

1 x1 x2 . . . xi−1 xi+1 . . . xn



 .

Old and New Inequalities

81

But this one is not very hard, because it follows immediately by multiplying the inequalities n−1  s  Y Y 1 1  1+ ≥ 1 + n−1 xj xj j6=i

j6=i

obtained from Huygens Inequality. Second solution: We will prove even more, that   2 n Y n  n Y 1 n −1 1 1+ ≥ · . x n 1 − xi i i=1 i=1 It is clear that this inequality is stronger than the initial one. First, let us prove that n  n Y 1 + xi n+1 . ≥ 1 − xi n−1 i=1 This follows from Jensen’s Inequality for the convex function f (x) = ln(1 + x) − ln(1 − x). So, it suffices to prove that n  n+1  2 n n Y n −1 n−1 2 · (1 − xi ) ≥ . n Y n i=1 xi i=1

But a quick look shows that this is exactly the inequality proved in the solution of the problem 121. 84. [ Vasile Cˆırtoaje, Gheorghe Eckstein ] Consider positive real numbers x1 , x2 , ..., xn such that x1 x2 ...xn = 1. Prove that 1 1 1 + + ... + ≤ 1. n − 1 + x1 n − 1 + x2 n − 1 + xn TST 1999, Romania First solution: Suppose the inequality is false for a certain system of n numbers. Then we can find a number k > 1 and n numbers which add up to 1, let them be ai , such that 1 = kai . Then we have n − 1 + xi  Y  n  n  Y 1 1 1= −n+1 < −n+1 . kai ai i=1 i=1

82

Solutions

We have used here the fact that ai <

1 . Now, we write 1 − (n − 1)ak = bk and n−1

we find that

n X

bk = 1 and also

k=1

n Y

(1 − bk ) < (n − 1)n b1 . . . bn . But this contradicts

k=1

the fact that for each j we have p 1 − bj = b1 + · · · + bj−1 + bj+1 + · · · + bn ≥ (n − 1) n−1 b1 . . . bj−1 bj+1 . . . bn . Second solution: Let us write the inequality in the form x1 x2 xn + + ... + ≥ 1. n − 1 + x1 n − 1 + x2 n − 1 + xn This inequality follows by summing the following inequalities 1− 1

1− 1

xn x1 x1 n xn n , . . . , , ≥ 1− 1 ≥ 1− 1 1− 1 1− 1 1− 1 1− 1 n − 1 + x1 n − 1 + xn x1 n + x2 n + . . . + xn n x1 n + x2 n + . . . + xn n The first from these inequalities is equivalent to 1 1− n

x1

1 1− n

+ x2

1 1− n

+ . . . + xn

1 −n

≥ (n − 1)x1

and follows from the AM-GM Inequality. Remark. 1 1 1 Replacing the numbers x1 , x2 , . . . , xn with , ,..., respectively, the inx1 x2 xn equality becomes as follows 1 1 1 + + ... + ≥ 1. 1 + (n − 1)x1 1 + (n − 1)x2 1 + (n − 1)xn 85. [ Titu Andreescu ] Prove that for any nonnegative real numbers a, b, c such that a2 + b2 + c2 + abc = 4 we have 0 ≤ ab + bc + ca − abc ≤ 2. USAMO, 2001 First solution (by Richard Stong): √ 3 The lower bound is not difficult. Indeed, we have ab + bc + ca ≥ 3 a2 b2 c2 and √ 3 thus it is enough to prove that abc ≤ 3 a2 b2 c2 , which follows from the fact that abc ≤ 4. The upper bound instead is hard. Let us first observe that there are two numbers among the three ones, which are both greater or equal than 1 or smaller than or equal to 1. Let them be b and c. Then we have 4 ≥ 2bc + a2 + abc ⇒ (2 − a)(2 + a) ≥ bc(2 + a) ⇒ bc ≤ 2 − a. Thus, ab + bc + ca − abc ≤ ab + 2 − a + ac − abc and it is enough to prove the inequality ab + 2 − a + ac − abc ≤ 2 ⇔ b + c − bc ≤ 1 ⇔ (b − 1)(c − 1) ≥ 0, which is true due to our choice.

Old and New Inequalities

83

Second solution: We won’t prove again the lower part, since this is an easy problem. Let us concentrate on the upper bound. Let a ≥ b ≥ c and let a = x + y, b = x − y. The hypothesis becomes x2 (2 + c) + y 2 (2 − c) = 4 − c2 and we have to prove that 2+c 2 (x2 − y 2 )(1 − c) ≤ 2(1 − xc). Since y 2 = 2 + c − x , the problem asks to 2−c 2 2 4x − (4 − c ) (1 − c) ≤ 2(1 − cx). Of course, we have c ≤ 1 prove the inequality 2−c √ 2+c 2 2 and 0 ≤ y = 2 + c − 2−c x ⇒ x2 ≤ 2 − c ⇒ x ≤ 2 − c (we have used the fact that a ≥ b ⇒ y ≥ 0 and b ≥ 0 ⇒ x ≥ y ≥ 0). Now, consider the √ 4x2 − (4 − c2 ) function f : [0, 2 − c] → R, f (x) = 2(1 − cx) − (1 − c). We have 2−c √ 1−c f 0 (x) = −2c − 8x · ≤ 0 and thus f is decreasing and f (x) ≥ f ( 2 − c). So, we 2−c have to prove that √ √ √ √ f ( 2 − c) ≥ 0 ⇔ 2(1−c 2 − c) ≥ (2−c)(1−c) ⇔ 3 ≥ c+2 2 − c ⇔ (1− 2 − c)2 ≥ 0 clearly true. Thus, the problem is solved. 86. [ Titu Andreescu ] Prove that for any positive real numbers a, b, c the following inequality holds √ √ √ √ √ √ a+b+c √ 3 − abc ≤ max{( a − b)2 , ( b − c)2 , ( c − a)2 }. 3 TST 2000, USA Solution: A natural idea would be to assume the contrary, which means that √ a+b+c √ 3 − abc ≤ a + b − 2 ab 3 √ a+b+c √ 3 − abc ≤ b + c − 2 bc 3 √ a+b+c √ 3 − abc ≤ c + a − 2 ca. 3 Adding these inequalities, we find that √ √ √ √ 3 a + b + c − 3 abc > 2(a + b + c − ab − bc − ca). √ √ √ √ Now, we will prove that a + b + c − 3 3 abc ≤ 2(a + b + c − ab − bc − ca) and the problem will be solved. Since the above inequality is homogeneous, we may assume √ √ √ that abc = 1. Then, it becomes 2 ab + 2 bc + 2 ca − a − b − c ≤ 3. Now, using Schur’s inequality, we find that for any positive reals x,y,z we have: p 9xyz 2xy + 2yz + 2zx − x2 − y 2 − z 2 ≤ ≤ 3 3 x2 y 2 z 2 x+y+z √ √ √ All we have to do is take x = a, y = b, z = c in the above inequality.

84

Solutions

87. [ Kiran Kedlaya ] Let a,b,c be positive real numbers. Prove that r √ √ a + ab + 3 abc a+b a+b+c 3 ≤ a· · . 3 2 3 Solution (by Anh Cuong): We have that √ a+ Now we will prove that r a+

3

ab +

√ 3

r 3

abc ≤ a +

a+b √ 3 ab + abc ≤ 2

ab

r 3

a·

a+b √ 3 + abc. 2

a+b a+b+c · . 2 3

By the AM-GM Inequality, we have 3a 2a + 1+ 3a 2a a + b a + b+c, 1· · ≤ a+b a+b+c 3

r 3

3b 2+ 3b a + b+c, 1·1· ≤ a+b+c 3

r 3

3c 2b + 1+ 2b 3c a + b a + b+c. 1· · ≤ a+b a+b+c 3 Now, just add them up and we have the desired inequality. The equality occurs when a = b = c. r 3

88. Find the greatest constant k such that for any positive integer n which is not √ √ a square, |(1 + n) sin(π n)| > k. Vietnamese IMO Training Camp, 1995 Solution: π We will prove that is the best constant. We must clearly have k < 2
√ √ √

2 2 1 + i + 1 | sin π i + 1 | for all positive integers i. Because | sin π i2 + 1 | = π π π k √ √ √ √ sin , we deduce that ≥ sin > , from i + i2 + 1 i + i2 + 1 i + i2 + 1 1 + i2 + 1 π where it follows that k ≤ . Now, let us prove that this constant is good. Clearly, the 2 inequality can be written √
π √ . sin π · { n} > 2(1 + n) We have two cases

Old and New Inequalities

85

√ 1 i) The first case is when { n} ≤ . Of course, 2 √ √ √ { n} ≥ n − n − 1 = √

n+

1 √

n−1

,

x3 we find that 6   3  √ π π 1 π √ sin(π{ n}) ≥ sin √ − . √ ≥ √ √ √ 6 n−1+ n n−1+ n n−1+ n

and because sin x ≥ x −

Let us prove that the last quantity is at least 2+

π √ . This comes down to 2(1 + n)

√ π2 n− n−1 √ >
2 , √ √ 1+ n 3 n+ n−1

√

√ √ √ √ √ or 6( n + n − 1)2 + 3( n + n − 1) > π 2 (1 + n) and it is clear. √ √ 1 1 and let n = ii) The second case is when { n} > . Let x = 1 − { n} < 2 2 √ 1 k 2 + p, 1 ≤ p ≤ 2k. Because { n} > ⇒ p ≥ k + 1. Then it is easy to see that 2 1 √ and so it suffices to prove that x≥ k + 1 + k 2 + 2k sin

k+1+

π √

k2

Using again the inequality sin x ≥ x −

+ 2k

≥

π √ . 2(1 + k 2 + k)

x3 we infer that, 6

√ √ 2 k 2 + k − k 2 + 2k − k + 1 π2 √ > √
2 . 1 + k2 + k 3 1 + k + k 2 + 2k √ √ But from the Cauchy-Schwarz Inequality we have 2 k 2 + k − k 2 + 2k − k ≥ 0.  p √
2 π2  Because the inequality 1 + k + k 2 + 2k > 1 + k 2 + k holds, this case is 3 also solved.

89. [ Dung Tran Nam ] Let x, y, z > 0 such that (x + y + z)3 = 32xyz. Find the x4 + y 4 + z 4 minimum and maximum of . (x + y + z)4 Vietnam, 2004 First solution (by Tran Nam Dung): We may of course assume that x + y + z = 4 and xyz = 2. Thus, we have to find the

86

Solutions

x4 + y 4 + z 4 . Now, we have 44 X x4 + y 4 + z 4 = (x2 + y 2 + z 2 )2 − 2 x2 y 2 = X X = (16 − 2 xy)2 − 2( xy)2 + 4xyz(x + y + z) =

extremal values of

=

2a2 − 64a + 288,

2 8 where a = xy +yz +zx. Because y +z = 4−x and yz = , we must have (4−x)2 ≥ , x √ x √ which implies that 3 − 5 ≤ x ≤ 2. Due to symmetry, we have x, y, z ∈ [3 − 5, 2]. This means that (x − 2)(y − 2)(z − 2) ≤ 0 and also √ √ √ (x − 3 + 5)(y − 3 + 5)(z − 3 + 5) ≥ 0. Clearing paranthesis, we deduce that # √ 5 5−1 a ∈ 5, . 2 "

(a − 16)2 − 112 x4 + y 4 + z 4 = , we find that the extremal values of the 44 128 √ √ √ ! √ 1+ 5 1+ 5 383 − 165 5 9 expression are , , attained for the triples 3 − 5, , , 256 128 2 2 respectively (2, 1, 1). But because

Second solution: As in the above solution, we must find the extremal values of x2 + y 2 + z 2 when 1 x + y + z = 1, xyz = , because after that the extremal values of the expression 32 a x4 + y 4 + z 4 can be immediately found. Let us make the substitution x = , y = 4 b c a2 + b2 + 4c2 2 2 2 , z = , where abc = 1, a + b + 2c = 4. Then x + y + z = and so we 4 2 16 2 must find the extremal values of a2 + b2 + 4c2 . Now, a2 + b2 + 4c2 = (4 − 2c)2 − + 4c2 c 1 2 and the problem reduces to finding the maximum and minimum of 4c − 8c − where c there are positive numbers a, b, c such that " abc√= 1,#a + b + 2c = 4. Of course, this 4 3− 5 comes down to (4 − 2c)2 ≥ , or to c ∈ , 1 . But this reduces to the study c 2 " # √ 1 3− 5 2 of the function f (x) = 4x − 8x − defined for , 1 , which is an easy task. x 2 90. [ George Tsintifas ] Prove that for any a, b, c, d > 0, (a + b)3 (b + c)3 (c + d)3 (d + a)3 ≥ 16a2 b2 c2 d2 (a + b + c + d)4 . Crux Mathematicorum

Old and New Inequalities

87

Solution: Let us apply Mac-Laurin Inequality for x = abc, y = bcd, z = cda, t = dab. We will find that X 

abc

4

3  ≥

X

a2 b2 c2 d2

abc · bcd · cda 4

=

4

X

a .

Thus, it is enough to prove the stronger inequality (a + b)(b + c)(c + d)(d + a) ≥ (a + b + c + d)(abc + bcd + cda + dab). Now, let us observe that (a + b)(b + c)(c + d)(d + a) = (ac + bd + ad + bc)(ac + bd + ab + cd) = P P = (ac + bd)2 + a2 (bc + bd + cd) ≥ 4abcd + a2 (bc + bd + cd) = = (a + b + c + d)(abc + bcd + cda + dab). And so the problem is solved. 91. [ Titu Andreescu, Gabriel Dospinescu ] Find the maximum value of the expression (bc)n (ca)n (ab)n + + 1 − ab 1 − bc 1 − ca where a, b, c are nonnegative real numbers which add up to 1 and n is some positive integer. Solution: First, we will treat the case n > 1. We will prove that the maximum value is 1 1 . It is clear that ab, bc, ca ≤ and so 3 · 4n−1 4 (ab)n (bc)n (ca)n 4 + + ≤ ((ab)n + (bc)n + (ca)n ) . 1 − ab 1 − bc 1 − ca 3 1 Thus, we have to prove that (ab)n + (bc)n + (ca)n ≤ n . Let a the maximum among 4 a, b, c. Then we have 1 ≥ an (1 − a)n = an (b + c)n ≥ an bn + an cn + nan bn−1 c ≥ an bn + bn cn + cn an . 4n 1 So, we have proved that in this case the maximum is at most . But for 3 · 4n−1 1 a = b = , c = 0 this value is attained and this shows that the maximum value is 2 1 for n > 1. Now, suppose that n = 1. In this case we have 3 · 4n−1 X ab X 1 = − 3. 1 − ab 1 − ab

88

Solutions

Using the fact that a + b + c = 1, it is not difficult to prove that X 3−2 ab + abc 1 1 1 X + + = . 1 − ab 1 − bc 1 − ca 1− ab + abc − a2 b2 c2 3 ab ≤ . With the above observations, this reduces to 1 − ab 8 X 3 − 27(abc)2 + 19abc ab ≤ . 11 But using Schur’s Inequality we infer that X 1 + 9abc ab ≤ 4 and so it is enough to show that We will prove that

X

3 − 27(abc)2 + 19abc 9abc + 1 ≤ ⇔ 108(abc)2 + 23abc ≤ 1, 4 11 1 which is true because abc ≤ . 27 3 Hence for n = 1, the maximum value is attained for a = b = c. 8 92. Let a, b, c be positive real numbers. Prove that 1 1 1 3 √ + + ≥ √ . 3 a(1 + b) b(1 + c) c(1 + a) abc(1 + 3 abc) Solution: The following observation is crucial   1 1 1 (1 + abc) + + +3 a(1 + b) b(1 + c) c(1 + a)

=

X 1 + abc + a + ab

= a(1 + b) X 1+a X b(c + 1) = + . a(1 + b) 1+b

We use now twice the AM-GM Inequality to find that X 1+a X b(c + 1) √ 3 3 + ≥ √ + 3 abc. 3 a(1 + b) 1+b abc And so we are left with the inequality √ 3 3 √ + 3 abc − 3 3 3 abc √ , ≥ √ 3 1 + abc abc(1 + 3 abc) which is in fact an identity! 93. [ Dung Tran Nam ] Prove that for any real numbers a, b, c such that a2 + b2 + c2 = 9, 2(a + b + c) − abc ≤ 10. Vietnam, 2002

Old and New Inequalities

89

First solution (by Gheorghe Eckstein): Because max{a, b, c} ≤ 3 and |abc| ≤ 10, it is enough to consider only the cases when a, b, c ≥ 0 or exactly of the three numbers is negative. First, we will suppose that a, b, c are nonnegative. If abc ≥ 1, then we are done, because p 2(a + b + c) − abc ≤ 2 3(a2 + b2 + c2 ) − 1 < 10. Otherwise, we may assume that a < 1. In this case we have ! r p b2 + c2 = 2a + 2 18 − 2a2 ≤ 10. 2(a + b + c) − abc ≤ 2 a + 2 2 Now, assume that not all three numbers are nonnegative and let c < 0. Thus, the problem reduces to proving that for any nonnegative x, y, z whose sum of squares is 9 we have 4(x + y − z) + 2xyz ≤ 20. But we can write this as (x−2)2 +(y−2)2 +(z−1)2 ≥ 2xyz − 6z − 2. Because 2xyz−6z−2 ≤ x(y 2 +z 2 )−6z−2 = −z 3 + 3z − 2 = −(z − 1)2 (z + 2) ≤ 0, the inequality follows. Second solution: √ Of course, we have |a|, |b|, |c| ≤ 3 and |a + b + c|, |abc| ≤ 3 3. Also, we may assume of course that a, b, c are non-zero and that a ≤ b ≤ c. If c < 0 then we have √ 2(a+b+c)−abc < −abc ≤ 3 3 < 10. Also, if a ≤ b < 0 < c then we have 2(a+b+c) < 2c ≤ 6 < 10 + abc because abc > 0. If a < 0 < b ≤ c, using the Cauchy-Schwarz Inequality we find that 2b + 2c − a ≤ 9. Thus, 2(a + b + c) = 2b + 2c − a + 3a ≤ 9 + 3a and it remains to prove that 3a − 1 ≤ abc. But a < 0 and 2bc ≤ 9 − a2 , so that it 9a − a3 remains to show that ≥ 3a − 1 ⇔ (a + 1)2 (a − 2) ≤ 0, which follows. So, we 2 just have to threat the case 0 < a ≤ b ≤ c. In this case we have 2b + 2c + a ≤ 9 and 2(a + b + c) ≤ 9 + a. So, we need to prove that a ≤ 1 + abc. This is clear if a < 1 and if a > 1 we have b, c > 1 and the inequality is again. Thus, the problem is solved.



94. [ Vasile Cˆırtoaje ] Let a, b, c be positive real numbers. Prove that         1 1 1 1 1 1 b+ −1 + b+ −1 c+ −1 + c+ −1 a + − 1 ≥ 3. a+ −1 b c c a a b

First solution: 1 1 1 With the notations x = a + − 1, y = b + − 1, z = c + − 1, the inequality b c a becomes xy + yz + zx ≥ 3. We consider without loss of generality that x = max{x, y, z}. From (x+1)(y+1)(z+1) = abc+

1 1 1 1 1 1 1 +a+b+c+ + + ≥ 2+a+b+c+ + + = 5+x+y+z, abc a b c a b c

90

Solutions

we get xyz + xy + yz + zx ≥ 4, with equality if and only if abc = 1. Because y+z =

1 (c − 1)2 +b+ > 0 we distinguish a c

two cases a) x> 0, yz ≤ 0; b)x> 0,y> 0,z> 0. a) x> 0, yz ≤ 0. We have xyz ≤ 0 and from xyz + xy + yz + zx ≥ 4 we get xy + yz + zx ≥ 4 > 3. b) x> 0,y> 0,z> 0. We denote xy + yz + zx = 3d2 with d > 0. From the mean inequalities we have p xy + yz + zx ≥ 3 3 x2 y 2 z 2 , from which we deduce that xyz ≤ d3 . On the basis of this result, from the inequality xyz + xy + yz + zx ≥ 4, we obtain d3 + 3d2 ≥ 4, (d − 1)(d + 2)2 ≥ 0, d ≥ 1 so xy + yz + zx ≥ 3. With this the given inequality is proved. We have equality in the case a = b = c = 1. Second solution: Let u = x + 1, v = y + 1, w = z + 1. Then we have uvw = u + v + w + abc +

1 ≥ u + v + w + 2. abc

Now, consider the function f (t) = 2t3 +t2 (u+v+w)−uvw. Because lim f (t) = ∞ and t→∞

f (1) ≤ 0, we can find a real number r ≥ 1 such that f (r) = 0. Consider the numbers v w u m = , n = , p = . They verify mnp = m + n + p + 2 we deduce from problem 49 r r r that mn + np + pm ≥ 2(m + n + p) ⇒ uv + vw + wu ≥ 2r(u + v + w) ≥ 2(u + v + w). But because u = x + 1, v = y + 1, w = z + 1, this last relation is equivalent to xy + yz + zx ≥ 3, which is what we wanted. 95. [ Gabriel Dospinescu ] Let n be an integer greater than 2. Find the greatest real number mn and the least real number Mn such that for any positive real numbers x1 , x2 , . . . , xn (with xn = x0 , xn+1 = x1 ), mn ≤

n X i=1

Solution: We will prove that mn = n X i=1

xi ≤ Mn . xi−1 + 2(n − 1)xi + xi+1

1 1 , Mn = . First, let us see that the inequality 2(n − 1) 2

xi 1 ≥ xi−1 + 2(n − 1)xi + xi+1 2(n − 1)

Old and New Inequalities

is trivial, because xi−1 + 2(n − 1)xi + xi+1 ≤ 2(n − 1) ·

91 n X

xk for all i. This shows

k=1

that mn ≥

1 . Taking xi = xi , the expression becomes 2(n − 1)

xn−1 1 (n − 2)x + + x + xn−1 + 2(n − 1) 1 + 2(n − 1)x + x2 1 + 2(n − 1)xn−1 + xn−2 and taking the limit when x approaches 0, we find that mn ≤ mn =

1 and thus 2(n − 1)

1 . 2(n − 1)

1 Now, we will prove that Mn ≥ . Of course, it suffices to prove that for any 2 x1 , x2 , . . . , xn > 0 we have n X i=1

xi 1 ≤ . xi−1 + 2(n − 1)xi + xi+1 2

But it is clear that n X i=1

n

X 2xi 2xi ≤ = √ xi−1 + 2(n − 1)xi + xi+1 2 xi−1 · xi+1 + 2(n − 1)xi i=1 =

n X i=1

√ Taking

xi−1 · xi+1 xi

1 √ . xi−1 · xi+1 n−1+ xi

= ai , we have to prove that if

n X

n Y

ai = 1 then

i=1

1 ≤ 1. But this has already been proved in the problem 84. Thus, n − 1 + ai i=1 1 and because for x1 = x2 = · · · = xn we have equality, we deduce that Mn ≥ 2 1 Mn = , which solves the problem. 2 96. [ Vasile Cˆırtoaje ] If x, y, z are positive real numbers, then x2

1 1 1 9 + 2 + 2 ≥ . 2 2 2 + xy + y y + yz + z z + zx + z (x + y + z)2 Gazeta Matematic˘ a

Solution: Considering the relation x2 + xy + y 2 = (x + y + z)2 − (xy + yz + zx) − (x + y + z)z,

92

Solutions

we get (x + y + z)2 1 = , z xy + yz + zx x2 + xy + y 2 − 1− 2 (x + y + z) x+y+z or

1 (x + y + z)2 = , x2 + xy + y 2 1 − (ab + bc + ca) − c x y z where a = ,b= ,c= . The inequality can be rewritten x+y+z x+y+z x+y+z as 1 1 1 + + ≥ 9, 1−d−c 1−d−b 1−d−a where a, b, c are positive reals with a + b + c = 1 and d = ab + bc + ca. After making some computations the inequality becomes 9d3 − 6d2 − 3d + 1 + 9abc ≥ 0 or d(3d − 1)2 + (1 − 4d + 9abc) ≥ 0. which is Schur’s Inequality. 97. [ Vasile Cˆırtoaje ] For any a, b, c, d > 0 prove that 2(a3 + 1)(b3 + 1)(c3 + 1)(d3 + 1) ≥ (1 + abcd)(1 + a2 )(1 + b2 )(1 + c2 )(1 + d2 ). Gazeta Matematic˘ a Solution: Using Huygens Inequality Y (1 + a4 ) ≥ (1 + abcd)4 , we notice that it is enough to show that that Y Y 24 (a3 + 1)4 ≥ (1 + a4 )(1 + a2 )4 . Of course, it suffices to prove that 2(a3 + 1)4 ≥ (a4 + 1)(a2 + 1)4 for any positive real a. But (a2 +1)4 ≤ (a+1)2 (a3 +1)2 and we are left with the inequality 2(a3 +1)2 ≥ (a + 1)2 (a4 + 1) ⇔ 2(a2 − a + 1)2 ≥ a4 + 1 ⇔ (a − 1)4 ≥ 0, which follows. 98. Prove that for any real numbers a, b, c, (a + b)4 + (b + c)4 + (c + a)4 ≥

4 4 (a + b4 + c4 ). 7 Vietnam TST, 1996

Old and New Inequalities

93

Solution: Let us Xmake the substitution X a + b = 2z, b + c = 2x, c + a = 2y. The inequality becomes (y + z − x)4 ≤ 28 x4 . Now, we have the following chain of identities 2  X 2 X  X X (y + z − x)4 = x2 + 2yz − 2xy − 2xz = 3 x2 + 4 x2 X  X 2  X  X  X (yz − xy − xz) + 4 (xy + xz − yz)2 = 3 x2 − 4 xy x2 + X 2 X 2  X 4 X X X +16 x2 y 2 − 4 xy = 4 x2 + 16 x2 y 2 − x ≤ 28 x4 X

because

X

x2

2

≤3

X

x4 ,

X

x2 y 2 ≤

X

x4 .

99. Prove that if a, b, c are positive real numbers such that abc = 1, then 1 1 1 1 1 1 + + ≤ + + . 1+a+b 1+b+c 1+c+a 2+a 2+b 2+c Bulgaria, 1997 Solution: Let x = a + b + c and y = ab + bc + ca. Using brute-force, it is easy to see that x2 + 4x + y + 3 12 + 4x + y the left hand side is 2 , while the right hand side is . Now, x + 2x + y + xy 9 + 4x + 2y the inequality becomes x2 + 4x + y + 3 12 + 4x + y 2x + 3 − xy 3−y −1≤ −1 ⇔ 2 ≤ . x2 + 2x + y + xy 9 + 4x + 2y x + 2x + y + xy 9 + 4x + 2y For the last inequality, we clear denominators. Then using the inequalities x ≥ 3, y ≥ 3, x2 ≥ 3y, we have 5 2 x2 y x y ≥ 5x2 , ≥ y 2 , xy 2 ≥ 9x, 5xy ≥ 15x, xy ≥ 3y and x2 y ≥ 27. 3 3 Summing up these inequalities, the desired inequality follows. 1 2 3 100. [ Dung Tran Nam ] Find the minimum value of the expression + + a b c where a, b, c are positive real numbers such that 21ab + 2bc + 8ca ≤ 12. Vietnam, 2001 First solution (by Dung Tran Nam): 1 2 3 Let = x, = y, = z. Then it is easy to check that the condition of the problem a b c becomes 2xyz ≥ 2x + 4y + 7z. And we need to minimize x + y + z. But   2xy > 7 z(2xy − 7) ≥ 2x + 4y ⇒ 2x + 4y  z≥ 2xy − 7

94

Solutions

Now, we transform the expression so that after one application of the AM-GM 2x + 4y Inequality the numerator 3xy − 7 should vanish x + y + z ≥ x + y + = 2xy − 7 14 r 2x + 7 11 7 11 x x+ +y− + ≥ x+ + 2 1 + 2 . But, it is immediate to prove 2x 2x 2xy − 7 2x x 7 r 3+ 7 3 9 15 x that 2 1 + 2 ≥ and so x + y + z ≥ + x + ≥ . We have equality for x 2 2 x 2 15 5 , achieved for x = 3, y = , z = 2. Therefore, in the initial problem the answer is 2 2 1 4 3 a = ,b = ,c = . 3 5 2 Second solution: We use the same substitution and reduce the problem to finding the minimum value of x + y + z when 2xyz ≥ 2x + 4y + 7z. Applying the weighted the AM-GM Inequality we find that  x+y+z ≥ 1

1

5x 2 7

 25 (3y) 1

1

1 3



15z 4

4  15

.

7

And also 2x + 4y + 7z ≥ 10 5 · 12 3 · 5 15 · x 5 · y 3 · z 15 . This means that (x+ 225 +y + z)2 (2x + 4y + 7z) ≥ xyz. Because 2xyz ≥ 2x + 4y + 7z, we will have 2 (x + y + z)2 ≥ with equality for x = 3, y =

225 15 ⇒x+y+z ≥ 4 2

5 , z = 2. 2

101. [ Titu Andreescu, Gabriel Dospinescu ] Prove that for any x, y, z, a, b, c > 0 such that xy + yz + zx = 3, a b c (y + z) + (z + x) + (x + y) ≥ 3. b+c c+a a+b

Solution: We will prove the inequality p a b c (y + z) + (z + x) + (x + y) ≥ 3(xy + yz + zx) b+c c+a a+b for any a, b, c, x, y, z. Because the inequality is homogeneous in x, y, z we can assume that x + y + z = 1. But then we can apply the Cauchy-Schwarz Inequality so that

Old and New Inequalities

95

to obtain a x + b+c

p b c y+ z + 3(xy + yz + zx) ≤ c+a a+b s r q r q   2 qX X a 3 X 3 X 2 ≤ x + xy + xy ≤ b+c 4 4 s s X  a 2 3 qX X X  a 2 3 + x2 + 2 xy = + . ≤ b+c 2 b+c 2 s X  a 2 3 X a Thus, we are left with the inequality + ≤ . But this one b+c 2 b+c X 3 ab ≥ , which is trivial. is equivalent to (c + a)(c + b) 4 Remark. A stronger inequality is the following: Xp a b c (y + z) + (z + x) + (x + y) ≥ (x + y)(x + z) − (x + y + z), b+c c+a a+b which may be obtained by applying the Cauchy-Schwarz Inequality, as follows b c a (y + z) + (z + x) + (x + y) = b+c c+a a+b   √ √ y+z z+x x+y 1 √ (a+b+c) + + −2(x+y +z) ≥ ( y + z + z + x+ x + y)2 − b+c c+a a+b 2 Xp 2(x + y + z) = (x + y)(x + z) − (x + y + z). A good exercise for readers is to show that Xp p (x + y)(x + z) ≥ x + y + z + 3(xy + yz + zx).

102. Let a, b, c be positive real numbers. Prove that (b + c − a)2 (c + a − b)2 (a + b − c)2 3 + + ≥ . (b + c)2 + a2 (c + a)2 + b2 (a + b)2 + c2 5 Japan, 1997 First solution: c+a a+b b+c ,y = ,z = . The inequality can be written Let x = a b c X (x − 1)2 3 ≥ . x2 + 1 5 Using the Cauchy-Schwarz Inequality, we find that X (x − 1)2 x2

+1

≥

(x + y + z − 3)2 x2 + y 2 + z 2 + 3

96

Solutions

and so it is enough to prove that X X X (x + y + z − 3)2 3 ≥ ⇔ ( x)2 − 15 x+3 xy + 18 ≥ 0. 2 2 2 x +y +z +3 5 But from X Schur’s Inequality, after some computations, we deduce that X xy ≥ 2 x. Thus, we have X 2 X  2 X X X x − 15 x+3 xy + 18 ≥ x −9 x + 18 ≥ 0, X the last one being clearly true since x ≥ 6. Second solution: Of course, we may take a + b + c = 2. The inequality becomes X X 4(1 − a)2 3 27 1 ≥ ⇔ ≤ . 2 2 + 2(1 − a) 5 1 + (1 − a)2 10 But with the substitution 1 − a = x, 1 − b = y, 1 − c = z, the inequality reduces to that from problem 47. 103. [ Vasile Cˆırtoaje, Gabriel Dospinescu ] Prove that if a1 , a2 , . . . , an ≥ 0 then  n a1 + a2 + · · · + an−1 n n n − an a1 + a2 + · · · + an − na1 a2 . . . an ≥ (n − 1) n−1 where an is the least among the numbers a1 , a2 , . . . , an . Solution: Let ai − an = xi ≥ 0 for i ∈ {1, 2, . . . , n − 1}. Now, let us look at  n−1 n X a i   n n X Y  i=1  n  ai − n · ai − (n − 1)  − an   n−1  i=1 i=1 as a polynomial in a = an . It is in fact an +

n−1 X i=1

(a + xi )n − na

n−1 Y i=1

 (a + xi ) − (n − 1)

x1 + x2 + · · · + xn−1 n−1

n .

We will prove that the coefficient of ak is nonnegative for all k ∈ {0, 1, . . . , n − 1}, because clearly the degree of this polynomial is at most n − 1. For k = 0, this follows from the convexity of the function f (x) = xn  n−1 n X xi   n−1 X   i=1  xni ≥ (n − 1)  n−1 .   i=1

Old and New Inequalities

For k > 0, the coefficient of ak is   n−1 n X n−k x −n k i=1 i

X

97

xi1 xi2 . . . xin−k .

1≤i1 <···
Let us prove that this is nonnegative. From the AM-GM Inequality we have X xi1 xi2 . . . xin−k ≤ n 1≤i1
n n−k

X



1≤i1
 n−k n−k xn−k + x + · · · + x i1 i2 in−k =

  n−1 k n X n−k · x n−1 k i=1 i

  n−1 n X n−k x . This shows that each coefficient of the k i=1 i polynomial is nonnegative and so this polynomial takes nonnegative values when restricted to nonnegative numbers. which is clearly smaller than

104. [ Turkevici ] Prove that for all positive real numbers x, y, z, t, x4 + y 4 + z 4 + t4 + 2xyzt ≥ x2 y 2 + y 2 z 2 + z 2 t2 + t2 x2 + x2 z 2 + y 2 t2 . Kvant Solution: Clearly, it is enough to prove the inequality if xyzt = 1 and so the problem becomes If a, b, c, d have product 1, then a2 + b2 + c2 + d2 + 2 ≥ ab + bc + cd + da + ac + bd. √ Let d the minimum among a, b, c, d and let m = 3 abc. We will prove that a2 + b2 + c2 + d2 + 2 − (ab + bc + cd + da + ac + bd) ≥ d2 + 3m2 + 2 − (3m2 + 3md), which is in fact   √ 3 a2 + b2 + c2 − ab − bc − ca ≥ d a + b + c − 3 abc . Because d ≤

√ 3

abc, proving this first inequality comes down to the inequality   √ √ 3 3 a2 + b2 + c2 − ab − bc − ca ≥ abc a + b + c − 3 abc .

a b c Take u = √ ,v = √ ,w = √ . Using problem 74, we find that 3 3 3 abc abc abc u2 + v 2 + w2 + 3 ≥ u + v + w + uv + vw + wu  √  √ which is exactly a2 + b2 + c2 − ab − bc − ca ≥ 3 abc a + b + c − 3 3 abc . Thus, it √ 3 remains to prove that d2 + 2 ≥ 3md ⇔ d2 + 2 ≥ 3 d2 , which is clear.

98

Solutions

105. Prove that for any real numbers a1 , a2 , . . . , an the following inequality holds !2 n n X X ij ai ≤ ai aj . i+j−1 i=1 i,j=1 Solution: Observe that n X

ij ai aj i + j−1 i,j=1

=

n X

1

=



n X

 0

= 0

 iai · jaj · ti−1+j−1  dt =

i,j=1 1

Z

ti+j−2 dt =

0

i,j=1

Z

1

Z iai · jaj

n X

!2 iai · t

i−1

.

i=1

Now, using the Cauchy-Schwarz Inequality for integrals, we get !2 ! !2 !2 Z 1 X Z 1 X n n n X i−1 i−1 iai · t dt ≥ iai · t dt = ai , 0

0

i=1

i=1

i=1

which ends the proof. 106. Prove that if a1 , a2 , . . . , an , b1 , . . . , bn are real numbers between 1001 and 2002, inclusively, such that a21 + a22 + · · · + a2n = b21 + b22 + · · · + b2n , then we have the inequality
a3 a3 17 2 a31 + 2 + ··· + n ≤ a1 + a22 + · · · + a2n . b1 b2 bn 10 TST Singapore Solution: The key ideas are that the inequality x2 + 1 ≤

    ai 1 1 ∈ , 2 for any i and that for all x ∈ , 2 we have bi 2 2

5 x. Consequently, we have 2 a2 5 5 ai · ≥ 1 + 21 ⇒ ai bi ≥ a2i + b2i 2 bi bi 2

and also n n n X X
5X ai bi ≥ a2i + b2i = 2 a2i (1) 2 i=1 i=1 i=1

Old and New Inequalities

Now, the observation that write

a2i b2i

99

a3i 5 ai a2 bi = and the inequality · ≥ 1 + 2i allow us to ai · bi 2 bi bi

5 2 a3 ai ≥ i + ai bi and adding up these inequalities yields 2 bi  X n n  n n X 5 X 2 X a3i a3i a ≥ + ai bi = + ai bi (2) 2 i=1 i bi b i=1 i=1 i i=1

Using (1) and (2) we find that
a3 a3 17 2 a31 + 2 + ··· + n ≤ a1 + a22 + · · · + a2n , b1 b2 bn 10 which is the desired inequality. 107. [ Titu Andreescu, Gabriel Dospinescu ] Prove that if a, b, c are positive real numbers which add up to 1, then (a2 + b2 )(b2 + c2 )(c2 + a2 ) ≥ 8(a2 b2 + b2 c2 + c2 a2 )2 .

Solution: 1 1 1 1 1 1 Let x = , y = , z = . We find the equivalent form if + + = 1 then a b c x y z (x2 + y 2 )(y 2 + z 2 )(z 2 + x2 ) ≥ 8(x2 + y 2 + z 2 )2 . We will prove the following inequality (x2 + y 2 )(y 2 + z 2 )(z 2 + x2 )



1 1 1 + + x y z

2

≥ 8(x2 + y 2 + z 2 )2

for any positive numbers x, y, z. Write x2 + y 2 = 2c, y 2 + z 2 = 2a, z 2 + x2 = 2b. Then the inequality becomes r X X abc ≥ a. b+c−a Recall Schur’s Inequality X X X a4 + abc(a + b + c) ≥ a3 (b + c) ⇔ abc(a + b + c) ≥ a3 (b + c − a). Now, using H¨ older’s Inequality, we find that X

a3 (b + c − a) =

3

a

X

X 3 a

 2 ≥ X 2 . 1 1 √ b+c−a b+c−a Combining the two inequalities, we find that r X X abc ≥ a b+c−a and so the inequality is proved. 

√

100

Solutions

108. [ Vasile Cˆırtoaje ] If a, b, c, d are positive real numbers such that abcd = 1, then 1 1 1 1 + + + ≥ 1. (1 + a)2 (1 + b)2 (1 + c)2 (1 + d)2 Gazeta Matematic˘ a Solution: If follows by summing the inequalities 1 1 1 + ≥ , (1 + a)2 (1 + b)2 1 + ab 1 1 1 + ≥ . (1 + c)2 (1 + d)2 1 + cd The first from these inequalities follows from 1 1 1 + − 2 2 (1 + a) (1 + b) 1 + ab

=

ab(a2 + b2 ) − a2 b2 − 2ab + 1 = (1 + a)2 (1 + b)2 (1 + c)2

=

ab(a − b)2 + (ab − 1)2 ≥ 0. (1 + a)2 (1 + b)2 (1 + ab)

Equality holds if a = b = c = d = 1. 109. [ Vasile Cˆırtoaje ] Let a, b, c be positive real numbers. Prove that a2 b2 c2 a b c + + ≥ + + . b2 + c2 c2 + a2 a2 + b2 b+c c+a a+b Gazeta Matematic˘ a Solution: We have the following identities b2

a ab(a − b) + ac(a − c) a2 − = + c2 b+c (b + c)(b2 + c2 )

b2 b bc(b − c) + ab(b − a) − = 2 2 c +a c+a (c + a)(c2 + a2 ) 2 c ac(c − a) + bc(c − b) c = . − a2 + b2 a+b (b + a)(b2 + a2 ) Thus, we have X a2

 X  ab(a − b) a ab(a − b) = − = b2 + c2 b+c (b + c)(b2 − c2 ) (a + c)(a2 + c2 )
X ab(a − b)2 ≥ 0. = a2 + b2 + c2 + ab + bc + ca · (b + c)(c + a)(b2 + c2 )(c2 + a2 ) −

X

Old and New Inequalities

101

110. [ Gabriel Dospinescu ] Let a1 , a2 , . . . , an be real numbers and let S be a non-empty subset of {1, 2, . . . , n}. Prove that !2 X

X

≤

ai

i∈S

(ai + . . . + aj )2 .

1≤i≤j≤n

TST 2004, Romania First solution: Denote sk = a1 + · · · + ak for k = 1, n and also sn+1 = 0. Define now ( 1 if i ∈ S bi = 0 , otherwise Using Abel’s summation we find that X ai = a1 b1 + a2 b2 + · · · + an bn = i∈S

= s1 (b1 − b2 ) + s2 (b2 − b3 ) + · · · + sn−1 (bn−1 − bn ) + sn bn + sn+1 (−b1 ) Now, put b1 − b2 = x1 , b2 − b3 = x2 , . . . , bn−1 − bn = xn−1 , bn = xn , xn+1 = −b1 . So we have n+1 X X xi si ai = i=1

i∈S n+1 X

and also xi ∈ {−1, 0, 1}. Clearly,

xi = 0. On the other hand, using Lagrange

i=1

identity we find that X

2

(ai + · · · + aj )

n X

=

i=1

1≤i≤j≤n

X

s2i +

1≤i<j≤n

X

=

(sj − si )2 =

2

(sj − si ) = (n + 1)

n+1 X

s2i

−

i=1

1≤i<j≤n+1

So we need to prove that (n + 1)

n+1 X

s2i

≥

i=1

n+1 X

!2 si xi

+

i=1

n+1 X

!2 si

.

i=1

But it is clear that n+1 X i=1

!2 si xi

+

n+1 X

!2 si

=

i=1

n+1 X

s2i (1 + x2i )

i=1

+

2

X 1≤i<j≤n+1

si sj (xi xj + 1)

n+1 X i=1

!2 si

.

102

Solutions

Now, using the fact that 2si sj ≤ si 2 + s2j , 1 + xi xj ≥ 0, we can write X X 2 si sj (xi xj + 1) ≤ (s2i + s2j )(1 + xi xj ) = 1≤i<jn+1

1≤i<j≤n+1

(s21 + · · · + s2n+1 ) + s21 x1 (x2 + x3 + · · · + xn ) + · · · + s2n xn (x1 +

n

+ x2 + · · · + xn−1 ) = −s21 x21 − · · · − s2n x2n + n(s21 + · · · + s2n ). So, n+1 X

!2 si xi

n+1 X

+

i=1

= (n + 1)

n+1 X

!2

n+1 X

≤

si

i=1

s2k (x2k + 1) +

k=1

n+1 X

s2k (n − x2k ) =

k=1

s2k and we are done.

k+1

Second solution (by Andrei Negut¸): First, let us prove a lemma Lemma For any a1 , a2 , . . . , a2k+1 ∈ R we have the inequality !2 k X X 2 a2i+1 ≤ (ai + · · · + aj ) . i=0

1≤i≤j≤2k+1

Proof of the lemma Let us take sk = a1 + · · · + ak . We have k X

a2i+1 = s1 + s3 − s2 + · · · + s2k+1 − s2k

i=0

and so the left hand side in the lemma is 2k+1 X

s2i + 2

i=1

X

X

s2i+1 s2j+1 + 2

0≤i<j≤k

X

s2i s2j − 2

1≤i<j≤k

s2i+1 s2j

0≤i≤k 1≤j≤k

and the right hand side is just (2k + 1)

2k+1 X

s2i − 2

i=1

X

si sj .

1≤i<j≤2k+1

Thus, we need to prove that 2k

2k+1 X i=1

s2i ≥ 4

X

s2i+1 s2j+1 + 4

0≤i<j≤k

X

s2i s2j

1≤i<j≤k

and it comes by adding up the inequalities 2s2j+1 s2i+1 ≤ s22i+1 + s22j+1 , 2s2i s2j ≤ s22i + s22j .

Old and New Inequalities

103

Now, let us turn back to the solution of the problem. Let us call a succession of ai ’s a sequence and call a sequence that is missing from S a gap. We group the successive sequences from S and thus S will look like this S = {ai1 , ai1 +1 , . . . , ai1 +k1 , ai2 , ai2 +1 , . . . , ai2 +k2 , . . . , air , . . . , air +kr } where ij + kj < ij+1 − 1. Thus, we write S as a sequence, followed by a gap, followed by a sequence, then a gap and so on. Now, take s1 = ai1 +· · ·+ai1 +k1 , s2 = ai1 +k1 +1 + · · · + ai2 −1 , . . . , s2r−1 = air + · · · + air +kr Then, !2 X ai = (s1 + s3 + · · · + s2r−1 )2 ≤ i∈S

X

2

(si + · · · + sj )

X

≤

1≤i≤j≤2r−1

2

(ai + · · · + aj )

1≤i≤j≤n

the last inequality being clearly true, because the terms in the left hand side are among those from the right hand side. Third solution: We will prove the inequality using induction. For n = 2 and n = 3 it’s easy. Suppose the inequality is true for all k < n and let us prove it for n. If 1 ∈ / S, then we just apply the inductive step for the numbers a2 , . . . , an , because the right hand side doesn’t decrease. Now, suppose 1 ∈ S. If 2 ∈ S, then we apply the inductive step with the numbers a1 + a2 , a3 , . . . , an . Thus, we may assume that 2 ∈ / S. It is easy to 2 (a + b) ≥ 2ab and thus we have (a1 + a2 + · · · + ak )2 + see that (a + b + c)2 + c2 ≥ 2 (a2 + a3 + · · · + ak−1 )2 ≥ 2a1 ak (*). Also, the inductive step for a3 , . . . , an shows that  2 X X 2   ≤ (ai + · · · + aj ) . 3≤i≤j≤n

i∈S\{1}

So, it suffices to show that a21

+ 2a1

X i∈S\{1}

ai ≤

n X

(a1 + · · · + ai ) +

i=1

But this is clear from the fact that up the inequalities from (*).

2

a21

n X

2

(a2 + · · · + ai )

i=2

appears in the right hand side and by summing

111. [ Dung Tran Nam ] Let x1 , x2 . . . , x2004 be real numbers in the interval [−1, 1] such that x31 + x32 + . . . + x32004 = 0. Find the maximal value of the x1 + x2 + · · · + x2004 . Solution: 1 Let us take ai = x3i and the function f : [−1, 1] → R, f (x) = x 3 . We will prove first the following properties of f :

104

Solutions

1. f (x + y + 1) + f (−1) ≥ f (x) + f (y) if −1 < x, y < 0. 2. f is convex on [−1, 0] and concave on [0, 1]. 3. If x > 0 and y < 0 and x + y < 0 then f (x) + f (y) ≤ f (−1) + f (x + y + 1) and if x + y > 0 then f (x) + f (y) ≤ f (x + y). The proofs of these results are easy. Indeed, for the first one we make the substitution x = −a3 , y = −b3 and it comes down to 1 > a3 +b3 +(1−a−b)3 ⇔ 1 > (a+b)(a2 −ab+b2 )+1−3(a+b)+3(a+b)2 −(a+b)3 ⇔ ⇔ 3(a + b)(1 − a)(1 − b) > 0, which follows. The second statement is clear and the third one can be easily deduced in the same manner as 1. From these arguments we deduce that if (t1 , t2 , . . . , t2004 ) = t is the point where the maximum value of the function g : A = {x ∈ [−1, 1]2004 |x1 + · · · + xn = 0} → 2004 X → R, g(x1 , . . . , x2004 ) = f (xk ) (this maximum exists because this function is k=1

defined on a compact) is attained then we have that all positive components of t are equal to each other and all negative ones are −1. So, suppose we have k components equal to −1 and 2004−k components equal to a number a. Because t1 +tr 2 +· · ·+t2004 = k k and the value of g in this point is (2004−k) 3 −k. 0 we find that a = 2004 − k 2004 − k r k Thus we have to find the maximum value of (2004 − k) 3 − k, when k is in 2004 − k the set {0, 1, . . . , 2004}. A short analysis with derivatives shows that the maximum is √ √ 3 attained when k = 223 and so the maximum value is 3 223 · 17812 − 223. 112. [ Gabriel Dospinescu, C˘alin Popa ] Prove that if n ≥ 2 and a1 , a2 , . . . , an are real numbers with product 1, then √ 2n a21 + a22 + · · · + a2n − n ≥ · n n − 1(a1 + a2 + · · · + an − n). n−1 Solution: We will prove the inequality by induction. For n = 2 it is trivial. Now, suppose the inequality is true for n−1 numbers and let us prove it for n. First, it is easy to see that it is enough to prove it for a1 , . . . , an > 0 (otherwise we replace a1 , a2 , . . . , an with |a1 |, |a2 |, . . . , |an |, which have product 1. Yet, the right hand side increases). Now, let an the maximum number among a1 , a2 , . . . , an and let G the geometric mean of a1 , a2 , . . . , an−1 . First, we will prove that √ 2n · n n − 1(a1 + a2 + · · · + an − n) ≥ a21 + a22 + · · · + a2n − n − n−1 √ 2n ≥ a2n + (n − 1)G2 − n − · n n − 1(an + (n − 1)G − n) n−1 which is equivalent to q a21 + a22 + · · · + a2n−1 − (n − 1) n−1 a21 a22 . . . a2n−1 ≥

Old and New Inequalities

105

√
2n √ · n n − 1 a1 + a2 + · · · + an−1 − (n − 1) n−1 a1 a2 . . . an−1 . n−1 √ √ Because, n−1 a1 a2 . . . an−1 ≤ 1 and a1 +a2 +· · ·+an−1 −(n−1) n−1 a1 a2 . . . an−1 ≥ 0, it is enough to prove the inequality √ 2n a21 + · · · + a2n−1 − (n − 1)G2 ≥ · n n − 1 · G · (a1 + · · · + an−1 − (n − 1)G). n−1 a1 an−1 Now, we apply the inductive hypothesis for the numbers , . . . , which have G G product 1 and we infer that   √ a21 + · · · + a2n−1 2(n − 1) n−1 a1 + · · · + an−1 −n+1≥ n−2 · −n+1 G2 n−2 G ≥

and so it suffices to prove that √ 2(n − 1) n−1 2n √ · · n n − 1(a1 +· · ·+an−1 −(n−1)G), n − 2(a1 +· · ·+an−1 −(n−1)G) ≥ n−2 n−1 √ n 1 n−1 √ ≥ n−1 . This becomes which is the same as 1 + n(n − 2) n−2 n(n−1)  (n − 1)n−1 1 ≥ 1+ n(n − 2) (n − 2)n and it follows for n > 4 from  1+

1 n(n − 2)

n(n−1) >2

and    n−2   1 1 e 1 1 (n − 1)n−1 · 1+ · 1+ < 1+ < 2. = (n − 2)n n−2 n−2 n−2 n−2 n−2 For n = 3 and n = 4 it is easy to check. Thus, we have proved that √ 2n a21 + a22 + · · · + a2n − n − · n n − 1(a1 + a2 + · · · + an − n) ≥ n−1 √ 2n · n n − 1(an + (n − 1)G − n) ≥ a2n + (n − 1)G2 − n − n−1 and it is enough to prove that   √ n−1 2n n−1 n 2(n−1) n−1 x + −n≥ · n−1 x + −n x2 n−1 x 1 ). Let us consider the function G   √ n−1 n−1 2n n n−1 f (x) = x2(n−1) + · n − 1 x + − n . − n − x2 n−1 x

for all x ≥ 1 (we took x =

We have f 0 (x) = 2 ·

  √ xn − 1 (n − 1)(xn + 1) n · − n n − 1 ≥0 x2 x

106

Solutions

because n−1

x

s 1 1 1 1 n−1 + =x . + ··· + ≥nn + x (n − 1)x (n − 1)x (n − 1)n−1

Thus, f is increasing and so f (x) ≥ f (1) = 0. This proves the inequality. 113. [ Vasile Cˆırtoaje ] If a, b, c are positive real numbers, then r r r 2a 2b 2c + + ≤ 3. a+b b+c c+a Gazeta Matematic˘ a First solution:

r

With the notations x =

b ,y= a

r

c ,z= b

r

a the problem reduces to proving c

that xyz = 1 implies r

r r 2 2 2 + + ≤ 3. 2 2 1+x 1+y 1 + z2 We presume that x ≤ y ≤ z, which implies xy ≤ 1 and z ≥ 1. We have !2 r r     2 2 2 2 1 − x2 y 2 + ≤ 2 + = 4 1 + ≤ 1 + x2 1 + y2 1 + x2 1 + y2 (1 + x2 )(1 + y 2 )   8 8z 1 − x2 y 2 = = . 4 1+ (1 + xy)2 1 + xy z+1 so r r r 2 2 2z + ≤2 2 2 1+x 1+y z+1 and we need to prove that r r 2z 2 2 + ≤ 3. z+1 1 + z2 Because r 2 2 ≤ 1 + z2 1+z we only need to prove that r 2z 2 2 + ≤ 3. z+1 1+z This inequality is equivalent to p √ √ 1 + 3z − 2 2z(1 + z) ≥ 0, ( 2z − z + 1)2 ≥ 0, and we are done.

Old and New Inequalities

107

Second solution: Clearly, the problem asks to prove that if xyz = 1 then r X 2 ≤ 3. x+1 We have two cases. The first and easy one is when xy + yz + zx ≥ x + y + z. In this case we can apply the Cauchy-Schwarz Inequality to get r r X 2 X 2 ≤ 3 . x+1 x+1 But X 1 X 3 ≤ ⇔ 2(xy + x + y + 1) ≤ 3(2 + x + y + z + xy + yz + zx) ⇔ x+1 2 ⇔ x + y + z ≤ xy + yz + zx and so in this case the inequality is proved. The second case is when xy + yz + zx < x + y + z. Thus, (x − 1)(y − 1)(z − 1) = x + y + z − xy − yz − zx > 0 and so exactly two of the numbers x, y, z are smaller than 1, let them be x and y. So, we must prove that if x and y are smaller than 1, then r r r 2 2 2xy + + ≤ 3. x+1 y+1 xy + 1 Using the Cauchy-Schwarz Inequality, we get r r r r r 2 2 2xy 1 2xy 1 + + ≤2 + + x+1 y+1 xy + 1 x+1 y+1 xy + 1 and so it is enough to prove that this last quantity is at most 3. But this comes down to 1 2xy 1 1− + −1 x+1 y+1 xy + 1 r r 2· ≤ . 2xy 1 1 1+ 1+ + xy + 1 x+1 y+1 1 1 Because we have + ≥ 1, the left hand side is at most x+1 y+1 1 1 1 − xy + −1= x+1 y+1 (x + 1)(y + 1) and so we are left with the inequality r 1 − xy 2xy r   ⇔ xy + 1 + (xy + 1) ≤ ≤ xy + 1 2xy (xy + 1) 1 + xy + 1 p ≤ xy + 1 + x + y ⇔ x + y ≥ 2xy(xy + 1) p √ which follows from 2xy(xy + 1) ≤ 2 xy ≤ x + y. The problem is solved. 1 − xy (x + 1)(y + 1)

108

Solutions

114. Prove the following inequality for positive real numbers x, y, z   1 1 9 1 + + ≥ (xy + yz + zx) 2 2 2 (x + y) (y + z) (z + x) 4 Iran, 1996 First solution (by Iurie Boreico): With the substitution x + y = c, y + z = a, z + x = b, the inequality becomes after some easy computations  X 2 1 − 2 (a − b)2 ≥ 0. ab c Let a ≥ b ≥ c. If 2c2 ≥ ab, each term in the above expression is positive and we are done. So, let 2c2 < ab. First, we prove that 2b2 ≥ ac, 2a2 ≥ bc. Suppose that 2b2 < ac. Then (b + c)2 ≤ 2(b2 + c2 ) < a(b + c) and so b + c < a, false. Clearly, we can write the inequality like that       2 1 1 2 2 1 2 2 − − − (a − c) + (b − c) ≥ (a − b)2 . ac b2 bc a2 c2 ab We can immediately see that the inequality (a − c)2 ≥ (a − b)2 + (b − c)2 holds and thus if suffices to prove that     2 1 1 1 2 2 1 2 + − 2 − 2 (b − c)2 ≥ + − − (a − b)2 . ac bc a b b2 c2 ab ac  2 1 2 2 1 1 1 − < − But it is clear that 2 + 2 − and so the right hand side b c ab ac b c (a − b)2 (b − c)2 is at most . Also, it is easy to see that b2 c2 2 2 1 1 1 1 (a − b)2 + − 2− 2 ≥ + > , ac bc a b ac bc b2 c2 (a − b)2 (b − c)2 and this ends the which shows that the left hand side is at least b2 c2 solution. Second solution: Since the inequality is homogenous, we may assume that xy + yz + zx = 3. Also, we make the substitution x + y + z = 3a. From (x + y + z)2 ≥ 3(xy + yz + zx), we get a ≥ 1. Now we write the inequality as follows 1 1 1 3 + + ≥ , 2 2 2 (3a − z) (3a − y) (3a − x) 4 4[(xy + 3az)2 + (yz + 3ax)2 + (zx + 3ay)2 ] ≥ 3(9a − xyz)2 , 4(27a4 − 18a2 + 3 + 4axyz) ≥ (9a − xyz)2 , 3(12a2 − 1)(3a2 − 4) + xyz(34a − xyz) ≥ 0,

(1)

Old and New Inequalities

12(3a2 − 1)2 + 208a2 ≥ (17a − xyz)2 .

109

(2)

We have two cases. i) Case 3a2 − 4 ≥ 0. Since 34a − xyz =

1 [34(x + y + z)(xy + yz + zx) − 9xyz] > 0, 9

the inequality (1) is true. ii) Case 3a2 − 4 < 0. From Schur’s Inequality (x + y + z)3 − 4(x + y + z)(xy + yz + zx) + 9xyz ≥ 0, it follows that 3a3 − 4a + xyz ≥ 0. Thus, 12(3a2 − 1)2 + 208a2 − (17a − xyz)2 ≥ 12(3a2 − 1)2 + 208a2 − a2 (3a2 + 13)2 = = 3(4 − 11a2 + 10a4 − 3a6 ) = 3(1 − a2 )2 (4 − 3a2 )2 ≥ 0. 115. Prove that for any x, y in the interval [0, 1], p p p √ 1 + x2 + 1 + y 2 + (1 − x)2 + (1 − y)2 ≥ (1 + 5)(1 − xy).

Solution (by Faruk F. Abi-Khuzam and Roy Barbara - ”A sharp inequality and the iradius conjecture”): p √ Let the function F : [0, 1]2 → R, F (x, y) = 1 + x2 + 1 + y2 + p √ 2 2 (1 − x) + (1 − y) − (1 + 5)(1 − xy). It is clear that F is symmetric in x and √ y and also the convexity of the function x → 1 + x2 shows that F (x, 0) ≥ 0 for all x. Now, suppose we fix y and consider F as a function in x. It’s derivatives are √ x 1−x f 0 (x) = (1 + 5)y + √ −p 2 x +1 (1 − x)2 + (1 − y)2 and 1 (1 − y)2 f 00 (x) = p +q . 3 (1 + x2 )3 ((1 − x)2 + (1 − y)2 ) Thus, f is convex and its derivative is increasing. Now, let s √ √ 6 3 √ , c = 1 + 5. r = 1− (1 + 5)2 1 The first case we will discuss is y ≥ . It is easy to see that in this case we have cy > 4 1 p (the derivative of the function y → y 2 c2 (y 2 − 2y + 2) − 1 is positive) y 2 − 2y + 2 and so f 0 (0) > 0. Because f 0 is increasing, we have f 0 (x) > 0 and so f is increasing with f (0) = F (0, y) = F (y, 0) ≥ 0. Thus, in this case the inequality is proved.

110

Solutions

Due to case  1 and  to symmetry, it remains  to showthat the inequality holds in 1 1 1 the cases x ∈ 0, , y ∈ [0, r] and x ∈ r, , y ∈ r, . 4 4 4 In the first case we deduce immediately that √ x 1−x −p f 0 (x) ≤ r(1 + 5) + √ 2 1+x 1 + (x − 1)2     1 1 Thus, we have f 0 < 0, which shows that f is decreasing on 0, . Because from 4 4   1 the first case discussed we have f ≥ 0, we will have F (x, y) = f (x) ≥ 0 for all 4   1 points (x, y) with x ∈ 0, , y ∈ [0, r]. 4     1 1 , y ∈ r, . Let Now, let us discuss the most important case, when x ∈ r, 4 4 the points O(0, 0), A(1, 0), B(1, 1), C(0, 1), M (1, y), N (x, 1). The triangle OM N has perimeter p p p 1 − xy 1 + x2 + 1 + y 2 + (1 − x)2 + (1 − y)2 and area . 2 But it is trivial to show that in any triangle with perimeter P and area S we have the p p √ P2 inequality S ≤ √ . Thus, we find that 1 + x2 + 1 + y 2 + (1 − x)2 + (1 − y)2 ≥ 12 3 √ p √ √ 6 3 1 − xy ≥ (1+ 5)(1−xy) due to the fact that xy ≥ r2 . The proof is complete. 116. [ Suranyi ] Prove that for any positive real numbers a1 , a2 , . . . , an the following inequality holds (n−1)(an1 +an2 +· · ·+ann )+na1 a2 . . . an ≥ (a1 +a2 +· · ·+an )(an−1 +an−1 +· · ·+an−1 ). n 1 2 Miklos Schweitzer Competition Solution: Again, we will use induction to prove this inequality, but the proof of the inductive step will be again highly non-trivial. Indeed, suppose the inequality is true for n numbers and let us prove it for n + 1 numbers. Due to the symmetry and homogeneity of the inequality, it is enough to prove it under the conditions a1 ≥ a2 ≥ · · · ≥ an+1 and a1 + a2 + · · · + an = 1. We have to prove that ! n n n n X Y Y X n+1 n+1 n n n ai + nan+1 + nan+1 · ai + an+1 · ai − (1 + an+1 ) ai + an+1 ≥ 0. i=1

i=1

i=1

i=1

But from the inductive hypothesis we have (n − 1)(an1 + an2 + · · · + ann ) + na1 a2 . . . an ≥ an−1 + an−1 + · · · + an−1 n 1 2

Old and New Inequalities

and so nan+1

n Y

ai ≥ an+1

n X

i=1

an−1 i

111

n X

− (n − 1)an+1

i=1

ani .

i=1

Using this last inequality, it remains to prove that ! ! n n n n X X X X n an+1 − ani − an+1 n ani − ain−1 + i i=1

i=1 n Y

+an+1

i=1

i=1

! ai + (n − 1)ann+1 − an−1 n+1

≥ 0.

i=1

Now, we will break this inequality into n Y

an+1

! ai + (n −

1)ann+1

−

an−1 n+1

≥0

i=1

and n

n X

an+1 i

i=1

−

n X

! ani

− an+1

n

i=1

n X

ani

−

i=1

n X

! an−1 i

≥ 0.

i=1

Let us justify these two inequalities. The first one is pretty obvious n n Y Y ai + (n − 1)ann+1 − an−1 (ai − an+1 + an+1 ) + (n − 1)ann+1 − an−1 n+1 = n+1 ≥ i=1

i=1

≥ ann+1 + an−1 n+1 ·

n X

n−1 (ai − an+1 ) + (n − 1)ann+1 − an+1 = 0.

i=1

Now, let us prove the second inequality. It can be written as n

n X

an+1 − i

i=1

Because n

n X i=1

ani

−

n X

n X

ani ≥ an+1

n

i=1

n X

ani −

i=1

n X

! an−1 i

.

i=1

an−1 ≥ 0 (using Chebyshev’s Inequality) and an+1 ≤ i

i=1

1 , it n

is enough to prove that n

n X i=1

an+1 i

−

n X i=1

ani

1 ≥ n

n

n X

ani

i=1

−

n X

! an−1 i

.

i=1

1 ≥ ani for all i. Thus, the inductive step but this one follows, because nan+1 + an−1 i n i is proved. 117. Prove that for any x1 , x2 , . . . , xn > 0 with product 1, n X X (xi − xj )2 ≥ x2i − n. 1≤i<j≤n

i=1

A generalization of Turkevici’s inequality

112

Solutions

Solution: Of course, the inequality can be written in the following form ! !2 n n X X 2 n ≥ −(n − 1) xk + xk . k=1

k=1

We will prove this inequality by induction. The case n = 2 is trivial. Suppose the inequality is true for n − 1 and let us prove it for n. Let ! !2 n n X X 2 f (x1 , x2 , . . . , xn ) = −(n − 1) xk + xk k=1

√

k=1

x2 x3 . . . xn , where we have already chosen x1 = min{x1 , x2 , . . . , xn }. and let G = It is easy to see that the inequality f (x1 , x2 , . . . , xn ) ≤ f (x1 , G, G, . . . , G) is equivalent to !2 ! n n n X X X x2k − xk ≥ 2x1 xk − (n − 1)G . (n − 1) n−1

k=2

k=2 n X

Clearly, we have x1 ≤ G and

k=2

xk ≥ (n − 1)G, so it suffices to prove that

k=2

(n − 1)

n X

x2k

−

k=2

n X

!2 xk

≥ 2G

k=2

n X

! xk − (n − 1)G .

k=2

We will prove that this inequality holds for all x2 , . . . , xn > 0. Because the inequality is homogeneous, it is enough to prove it when G = 1. In this case, from the induction step we already have !2 n n X X (n − 2) x2k + n − 1 ≥ xk k=2

k=2

and so it suffices to prove that n X k=2

x2k + n − 1 ≥

n X k=2

2xk ⇔

n X

(xk − 1)2 ≥ 0,

k=2

clearly true. Thus, we have proved that f (x1 , x2 , . . . , xn ) ≤ f (x1 , G, G, . . . , G). Now, to complete the inductive step, we will prove that f (x1 , G, G, . . . , G) ≤ 0. Because clearly 1 x1 = n−1 , the last assertion reduces to proving that G    2 1 1 (n − 1) (n − 1)G2 + 2(n−1) + n ≥ (n − 1)G + n−1 G G which comes down to

2n − 2 n−2 + n ≥ n−2 2n−2 G G and this one is an immediate consequence of the AM-GM Inequality.

Old and New Inequalities

113

118. [ Gabriel Dospinescu ] Find the minimum value of the expression n r X a1 a2 . . . an 1 − (n − 1)ai i=1 where a1 , a2 , . . . , an <

1 add up to 1 and n > 2 is an integer. n−1

Solution: 1 We will prove that the minimal value is √ . Indeed, using Suranyi’s Innn−3 equality, we find that (n − 1)

n X

ani + na1 a2 . . . an ≥

i=1

n X

an−1 ⇒ na1 a2 . . . an ≥ i

i=1

n X

an−1 (1 − (n − 1)ai ). i

i=1

Now, let us observe that n X k=1

an−1 (1 − (n − 1)ak ) = k

3 q n−1 3 ak

n X k=1

r

1 1 − (n − 1)ak

2

and so, by an immediate application of H¨ older Inequality, we have !3 n X n−1 3 ak n X k=1 n−1 ai (1 − (n − 1)ai ) ≥ s !2 . n X i=1 1 1 − (n − 1)ak k=1

But for n > 3, we can apply Jensen’s Inequality to deduce that  n  n−1 n 3 X X n−1 3 ak a k   k=1  1 k=1  ≥ = n−1 .  n  n n 3   Thus, combining all these inequalities, we have proved the inequality for n > 3. For n = 3 it reduces to proving that r X X abc ≥ a b+c−a which was proved in the solution of the problem 107. 119. [ Vasile Cˆırtoaje ] Let a1 , a2 , . . . , an < 1 be nonnegative real numbers such that r √ a21 + a22 + . . . + a2n 3 a= ≥ . n 3

114

Solutions

Prove that

a1 a2 an na + + ... + ≥ . 2 2 2 1 − a1 1 − a2 1 − an 1 − a2

Solution: We proceed by induction. Clearly, the inequality is trivial for n= 1. Now we suppose that the inequality is valid for n = k − 1, k ≥ 2 and will prove that a2 ak ka a1 + + ... + ≥ , 1 − a21 1 − a22 1 − a2k 1 − a2 for

√ a21 + a22 + . . . + a2k 3 a= ≥ . k 3 We assume, without loss of generality, that a1 ≥ a2 ≥ . . . ≥ ak , therefore a ≥ ak . Using the notation s s a21 + a2 + · · · + a2k−1 ka2 − a2k x= = , k−1 k−1 √ 3 from a ≥ ak it follows that x ≥ a ≥ . Thus, by the inductive hypothesis, we have 3 a2 ak−1 (k − 1)x a1 + + ... + ≥ , 1 − a21 1 − a22 1 − a2k−1 1 − x2 r

and it remains to show that ak (k − 1)x ka + ≥ . 1 − a2k 1 − x2 1 − a2 From x2 − a2 =

ka2 − a2k a2 − a2k − a2 = , k−1 k−1

we obtain (k − 1)(x − a) =

(a − ak )(a + ak ) x+a

and     ak (k − 1)x ka ak a x a + − = − + (k − 1) − = 1 − a2k 1 − x2 1 − a2 1 − a2k 1 − a2 1 − x2 1 − a2   a − ak −(1 + aak ) (a + ak )(1 + ax) −(a − ak )(1 + aak ) (k − 1)(x − a)(1 + ax) + = + = (1 − a2k )(1 − a2 ) (1 − x2 )(1 − a2 ) 1 − a2 1 − a2k (1 − x2 )(x + a) =

(a − ak )(x − ak )[−1 + x2 + a2k + xak + a(x + ak ) + a2 + axak (x + ak ) + a2 xak ] . (1 − a2 )(1 − a2k )(1 − x2 )(x + a)

Since x2 − a2k =

k(a − ak )(a + ak ) ka2 − a2k − a2k = , it follows that k−1 k−1 (a − ak )(x − ak ) =

k(a − ak )2 (a + ak ) ≥ 0, (k − 1)(x + ak )

Old and New Inequalities

115

and hence we have to show that x2 + a2k + xak + a(x + ak ) + a2 + axak (x + ak ) + a2 xak ≥ 1. In order to show this inequality, we notice that x2 + a2k =

ka2 + (k − 2)a2k ka2 ≥ k−1 k−1

and r q k 2 2 x + ak ≥ x + ak ≥ a . k−1 Consequently, we have x2 + a2k + xak + a(x + ak ) + a2 + axak (x + ak ) + a2 xak ≥ (x2 + a2k ) + a(x + ak ) + a2 ≥ ! r k k ≥ + + 1 a2 > 3a2 = 1, k−1 k−1 and the proof is complete. Equality holds when a1 = a2 = . . . = an . Remarks. 1. From the final solution, we can easily see that the inequality is valid for the larger condition 1 a≥ r . q n n 1 + n−1 + n−1 This is the largest r range for a, because for a1 = a2 = · · · = an−1 = x and an = 0 n−1 1 . (therefore a = x ), from the given inequality we get a ≥ r q n n n + n−1 1 + n−1 √ 3 2. The special case n = 3 and a = is a problem from Crux 2003. 3 120. [ Vasile Cˆırtoaje, Mircea Lascu ] Let a, b, c, x, y, z be positive real numbers such that (a + b + c)(x + y + z) = (a2 + b2 + c2 )(x2 + y 2 + z 2 ) = 4. Prove that abcxyz <

1 . 36

Solution: Using the AM-GM Inequality, we have    4(ab+bc+ca)(xy+yz+zx) = (a + b + c)2 − (a2 + b2 + c2 ) (x + y + z)2 − (x2 + y 2 + z 2 ) = = 20 − (a + b + c)2 (x2 + y 2 + z 2 ) − (a2 + b2 + c2 )(x + y + z)2 ≤ p ≤ 20 − 2 (a + b + c)2 (x2 + y 2 + z 2 )(a2 + b2 + c2 )(x + y + z)2 = 4,

116

Solutions

therefore (ab + bc + ca)(xy + yz + za) ≤ 1.

(1)

By multiplying the well-known inequalities (ab + bc + ca)2 ≥ 3abc(a + b + c),

(xy + yz + zx)2 ≥ 3xyz(x + y + z),

it follows that (ab + bc + ca)2 (xy + yz + zx)2 ≥ 9abcxyz(a + b + c)(x + y + z), or (ab + bc + ca)(xy + yz + zx) ≥ 36abcxyz. (2) From (1) and (2), we conclude that 1 ≤ (ab + bc + ca)(xy + yz + zx) ≥ 36abcxyz, therefore 1 ≤ 36abcxyz. To have 1 = 36abcxyz, the equalities (ab + bc + ca)2 = 3abc(a + b + c) and (xy + yz + zx)2 = 3xyz(x + y + z) are necessary. But these conditions imply a = b = c and x = y = z, which contradict the relations (a + b + c)(x + y + z) = (a2 + b2 + c2 )(x2 + y 2 + z 2 ) = 4. Thus, it follows that 1 > 36abcxyz. 121. [ Gabriel Dospinescu ] For a given n > 2, find the minimal value of the constant kn , such that if x1 , x2 , . . . , xn > 0 have product 1, then √

1 1 1 +√ + ··· + √ ≤ n − 1. 1 + kn x1 1 + kn x2 1 + kn xn Mathlinks Contest

Solution: 2n − 1 We will prove that kn = . Taking x1 = x2 = · · · = xn = 1, we find that (n − 1)2 2n − 1 kn ≥ . So, it remains to show that (n − 1)2 n X

1 ≤n−1 2n − 1 1+ x k (n − 1)2

r k=1

if x1 · x2 · · · · · xn = 1. Suppose this inequality doesn’t hold for a certain system of n numbers with product 1, x1 , x2 , . . . , xn > 0. Thus, we can find a number M > n − 1 and some numbers ai > 0 which add up to 1, such that 1 = M ak . 2n − 1 x 1+ k (n − 1)2

r

Old and New Inequalities

117

1 and we have n−1   n Y   n  n  Y 2n − 1 1 1 (n − 1)2 < −1 ⇒ −1 . 1= 2n − 1 M 2 a2k (n − 1)2 (n − 1)2 a2k

Thus, ak <

k=1

k=1

Now, denote 1 − (n − 1)ak = bk > 0 and observe that

n X

bk = 1. Also, the above

k=1

inequality becomes (n − 1)2n

n Y

bk ·

k=1

n Y

(2 − bk ) > (2n − 1)n

k=1

n Y

!2 (1 − bk )

.

k=1

Because from the AM-GM Inequality we have n  n Y 2n − 1 , (2 − bk ) ≤ n k=1

our assumption leads to n Y

(1 − bk )2 <

k=1

(n − 1)2n b1 b2 . . . b n . nn

So, it is enough to prove that for any positive numbers a1 , a2 , . . . , an the inequality n Y

(a1 + a2 + · · · + ak−1 + ak+1 + · · · + an )2 ≥

k=1

(n − 1)2n a1 a2 . . . an (a1 + a2 + · · · + an )n nn

holds. This strong inequality will be proved by induction. For n = 3, it follows from the fact that   X   X  2 Y 2 8 · a · ab (a + b) 64 X 3 9 ≥ ≥ a . abc abc 27 Suppose the inequality is true for all systems of n numbers. Let a1 , a2 , . . . , an+1 be positive real numbers. Because the inequality is symmetric and homogeneous, we may assume that a1 ≤ a2 ≤ · · · ≤ an+1 and also that a1 + a2 + · · · + an = 1. Applying the inductive hypothesis we get the inequality n Y

(1 − ai )2 ≥

i=1

(n − 1)2n a1 a2 . . . an . nn

To prove the inductive step, we must prove that n Y i=1

(an+1 + 1 − ai )2 ≥

n2n+2 a1 a2 . . . an an+1 (1 + an+1 )n+1 . (n + 1)n+1

Thus, it is enough to prove the stronger inequality 2 n  Y an+1 n3n+2 a (1 + an+1 )n+1 . 1+ ≥ 2n n+1 n+1 1 − a (n − 1) · (n + 1) i i=1

118

Solutions

Now, using Huygens Inequality and the AM-GM Inequality, we find that 2n  n  Y i=1

1+

an+1 1 − ai

2

   ≥ 1 +  

   a  v n+1  u n  uY n t (1 − ai ) 

 ≥

1+

nan+1 n−1

2n

i=1

and so we are left with the inequality  2n n3n+2 nan+1 ≥ an+1 (1 + an+1 )n+1 1+ 2n n−1 (n − 1) · (n + 1)n+1 1 n(1 + an+1 ) . So, we can put = 1 + x, where x is n n+1 nonnegative. So, the inequality becomes 2n  1 + (n + 1)x x ≥ . 1+ n(x + 1) (x + 1)n−1 if an+1 ≥ max{a1 , a2 , . . . , an } ≥

Using Bernoulli Inequality, we find immediately that  2n x 3x + 1 . 1+ ≥ n(x + 1) x+1 Also, (1 + x)n−1 ≥ 1 + (n − 1)x and so it is enough to prove that 3x + 1 1 + (n + 1)x ≥ x+1 1 + (n − 1)x which is trivial. So, we have reached a contradiction assuming the inequality doesn’t hold for a certain system of n numbers with product 1, which shows that in fact the inequality 2n − 1 is true for kn = and that this is the value asked by the problem. (n − 1)2 Remark. For n = 3, we find an inequality stronger than a problem given in China Mathematical Olympiad in 2003. Also, the case n = 3 represents a problem proposed by Vasile Cˆ artoaje in Gazeta Matematic˘a, Seria A. 122. [ Vasile Cˆırtoaje, Gabriel Dospinescu ] For a given n > 2, find the maximal value of the constant kn such that for any x1 , x2 , . . . , xn > 0 for which x21 + x22 + · · · + x2n = 1 we have the inequality (1 − x1 )(1 − x2 ) . . . (1 − xn ) ≥ kn x1 x2 . . . xn .

Old and New Inequalities

119

Solution: √ n We will prove that this constant is ( n − 1) . Indeed, let ai = x2i . We must find the minimal value of the expression n Y √ (1 − ai ) i=1 n Y √

ai

i=1

√ when a1 +a2 +· · ·+an = 1. Let us observe that proving that this minimum is ( n−1)n reduces to proving that n n n Y Y Y √ √ √ (1 − ai ) ≥ ( n − 1)n · ai · (1 + ai ). i=1

i=1

i=1

But from the result proved in the solution of the problem 121, we find that  n Y n n Y (n − 1)2 (1 − ai )2 ≥ ai . n i=1 i=1 So, it is enough to prove that n Y i=1

(1 +

√

 ai ) ≤

1 1+ √ n

n .

But this is an easy task, because from the AM-GM Inequality we get  n n X √ ai    n n Y   √ 1 i=1   ≤ 1+ √ (1 + ai ) ≤ 1 + , n  n   i=1 the last one being a simple consequence of the Cauchy-Schwarz Inequality. Remark. The case n = 4 was proposed by Vasile Cˆartoaje in Gazeta Matematic˘a Annual Contest, 2001.

Glossary

(1) Abel’s Summation Formula If a1 , a2 , ..., an , b1 , b2 , ..., bn are real or complex numbers, and Si = a1 + a2 + ... + ai ,i = 1, 2, ..., n, then n X

ai bi =

i=1

n−1 X

Si (bi − bi+1 ) + Sn bn .

i=1

(2) AM-GM (Arithmetic Mean-Geometric Mean) Inequality If a1 , a2 , ..., an are nonnegative real numbers, then n

1 1X ai ≥ (a1 a2 ...an ) n , n i=1

with equality if and only if a1 = a2 = ... = an . This inequality is a special case of the Power Mean Inequality.

(3) Arithmetic Mean-Harmonic Mean (AM-HM) Inequality If a1 , a2 , ..., an are positive real numbers, then n

1X ai ≥ n i=1

1 n X 1 1 n a i=1 i

with equality if and only if a1 = a2 = ... = an . This inequality is a special case of the Power Mean Inequality.

(4) Bernoulli’s Inequality For any real numbers x > −1 and a > 1 we have (1 + x)a ≥ 1 + ax.

121

122

Solutions

(5) Cauchy-Schwarz’s Inequality For any real numbers a1 , a2 , ..., an and b1 , b2 , ..., bn (a21 + a22 + ... + a2n )(b21 + b22 + ... + b2n ) ≥ ≥ (a1 b1 + a2 b2 + ... + an bn )2 , with equality if and only if ai and bi are proportional, i = 1, 2, ..., n. (6) Cauchy-Schwarz’s Inequality for integrals If a, b are real numbers, a < b, and f, g : [a, b] → R are integrable functions, then !2 ! ! Z b Z b Z b f (x)g(x)dx ≤ f 2 (x)dx · g 2 (x)dx . a

a

a

(7) Chebyshev’s Inequality If a1 ≤ a2 ≤ . . . ≤ an and b1 , b2 , . . . , bn are real numbers, ! then ! n n n X X X 1 ai · bi ; ai bi ≥ 1)If b1 ≤ b2 ≤ . . . ≤ bn then n i=1 i=1 i=1 ! ! n n n X X 1 X 2)If b1 ≥ b2 ≥ . . . ≥ bn then ai bi ≤ ai · bi ; n i=1 i=1 i=1 (8) Chebyshev’s Inequality for integrals If a, b are real numbers, a < b, and f, g : [a, b] → R are integrable functions and having the same monotonicity, then Z b Z b Z b (b − a) f (x)g(x)dx ≥ f (x)dx · g(x)dx a

a

a

and if one is increasing, while the other is decreasing the reversed inequality is true. (9) Convex function A real-valued function f defined on an interval I of real numbers is convex if, for any x, y in I and any nonnegative numbers α, β with sum 1, f (αx + βy) ≤ αf (x) + βf (y). (10) Convexity A function f (x) is concave up (down) on [a, b] ⊆ R if f (x) lies under (over) the line connecting (a1 , f (a1 )) and (b1 , f (b1 )) for all a ≤ a1 < x < b1 ≤ b. A function g(x) is concave up (down) on the Euclidean plane if it is concave up (down) on each line in the plane, where we identify the line naturally

Old and New Inequalities

123

with R. Concave up and down functions are also called convex and concave, respectively. If f is concave up on an interval [a, b] and λ1 , λ2 , ..., λn are nonnegative numbers wit sum equal to 1, then λ1 f (x1 ) + λ2 f (x2 ) + ... + λn f (xn ) ≥ f (λ1 x1 + λ2 x2 + ... + λn xn ) for any x1 , x2 , ..., xn in the interval [a, b]. If the function is concave down, the inequality is reversed. This is Jensen’s Inequality. (11) Cyclic Sum Let n be a positive integer. Given a function f of n variables, define the cyclic sum of variables (x1 , x2 , ..., xn ) as X f (x1 , x3 , ..., xn ) = f (x1 , x2 , ..., xn ) + f (x2 , x3 , ..., xn , x1 ) cyc

+... + f (an , a1 , a2 , ..., an−1 ).

(12) H¨ older’s Inequality 1 1 If r, s are positive real numbers such that + = 1, then for any positive r s real numbers a1 , a2 , . . . , an , b1 , b 2 , . . . , b n , n X

 ai bi

i=1

n

n X

  i=1 ≤  n 

 r1  ari 

n X

   i=1  ·   n  

 1s bsi 

  .  

(13) Huygens Inequality If p1 , p2 , . . . , pn , a1 , a2 , . . . , an , b1 , b2 , . . . , bn are positive real numbers with p1 + p2 + . . . + pn = 1, then n Y i=1

(ai + bi )pi ≥

n Y i=1

api i +

n Y i=1

(14) Mac Laurin’s Inequality For any positive real numbers x1 , x2 , . . . , xn , S1 ≥ S2 ≥ . . . ≥ Sn ,

bpi i .

124

Solutions

where

v X u u xi1 · xi2 · . . . · xik u u k 1≤i1
(15) Minkowski’s Inequality For any real number r ≥ 1 and any positive real numbers a1 , a2 , . . . , an , b1 , b2 , . . . , bn , ! r1 ! r1 ! r1 n n n X X X r r r (ai + bi ) ≤ ai + bi . i=1

i=1

i=1

(16) Power Mean Inequality For any positive real numbers a1 , a2 , . . . , an with sum equal to 1, and any positive real numbers x1 , x2 , . . . , xn , we define Mr = (a1 xr1 + a2 xr2 + 1 . . . + an xrn ) r if r is a non-zero real and M0 = xa1 1 xa2 2 . . . xann , M∞ = max{x1 , x2 , . . . , xn }, M−∞ = min{x1 , x2 , . . . , xn }. Then for any reals s ≤ t we have M−∞ ≤ Ms ≤ Mt ≤ M∞ . (17) Root Mean Square Inequality If a1 , a2 , ..., an are nonnegative real numbers, then r a21 + a22 + ... + a2n a1 + a2 + ... + an ≥ , n n with equality if and only if a1 = a2 = ... = an . (18) Schur’s Inequality For any positive real numbers x, y, z and any r > 0, xr (x − y)(x − z) + y r (y − z)(y − x) + z r (z − x)(z − y) ≥ 0. The most common case is r = 1, which has the following equivalent forms: 1)

x3 + y 3 + z 3 + 3xyz ≥ xy(x + y) + yz(y + z) + zx(z + x);

2)

xyz ≥ (x + y − z)(y + z − z)(z + x − y);

3)

if

x+y+z =1

then

xy + yz + zx ≤

1 + 9xyz . 4

(19) Suranyi’s Inequality For any nonnegative real numbers a1 , a2 , . . . , an , ! ! n n n n X Y X X n−1 n (n − 1) ak + n ak ≥ ak · ak . k=1

k=1

k=1

k=1

Old and New Inequalities

125

(20) Turkevici’s Inequality For any positive real numbers x, y, z, t, x4 + y 4 + z 4 + t4 + 2xyzt ≥ x2 y 2 + y 2 z 2 + z 2 t2 + t2 x2 + x2 z 2 + y 2 t2 . (21) Weighted AM-GM Inequality For any nonnegative real numbers a1 , a2 , ..., an , if w1 , w2 , ..., wn are nonnegative real numbers (weights) with sum 1, then wn 1 w2 w1 a1 + w2 a2 + ... + wn an ≥ aw 1 a2 . . . an ,

with equality if and only if a1 = a2 = ... = an .

Further reading 1. Andreescu, T.; Feng, Z., 101 Problems in Algebra from the Training of the USA IMO Team, Australian Mathematics Trust, 2001. 2. Andreescu, T.; Feng, Z., USA and International Mathematical Olympiads 2000, 2001, 2002, 2003, MAA. 3. Andreescu, T.; Feng, Z., Mathematical Olympiads: Problems and Solutions from around the World, 1998-1999, 1999-2000, MAA. 4. Andreescu, T.; Kedlaya, K., Mathematical Contests 1995-1996, 1996-1997, 1997-1998: Olympiad Problems from around the World, with Solutions, AMC. 5. Andreescu, T.; Enescu, B., Mathematical Olympiad Treasures, Birkh¨auser, 2003. 6. Andreescu, T.; Gelca, R., Mathematical Olympiad Challenges, Birkh¨auser, 2000. 7. Andreescu, T.; Andrica, D., 360 Problems for Mathematical Contests, GIL Publishing House, 2002. 8. Becheanu, M., Enescu, B., Inegalitat¸i elementare. . . ¸si mai put¸in elementare, GIL Publishing House, 2002 9. Beckenbach, E., Bellman, R. Inequalities, Springer Verlag, Berlin, 1961 10. Drimbe, M.O., Inegalit˘ a¸ti idei ¸si metode, GIL Publishing House, 2003 11. Engel, A., Problem-Solving Strategies, Problem Books in Mathematics, Springer, 1998. 12. G.H. Littlewood, J.E. Polya, G., Inequalities, Cambridge University Press, 1967 13. Klamkin, M., International Mathematical Olympiads, 1978-1985, New Mathematical Library, Vol. 31, MAA, 1986. 14. Larson, L. C., Problem-Solving Through Problems, Springer-Verlag, 1983. 15. Lascu, M., Inegalit˘ a¸ti, GIL Publishing House, 1994 16. Liu, A., Hungarian Problem Book III, New Mathematical Library, Vol. 42, MAA, 2001. 17. Lozansky, E; Rousseau, C., Winning Solutions, Springer, 1996. 18. Mitrinovic, D.S., Analytic inequalities, Springer Verlag, 1970 19. Panaitopol, L., B˘ andil˘ a, V., Lascu, M., Inegalit˘ a¸ti, GIL Publishing House, 1995 20. Savchev, S.; Andreescu, T., Mathematical Miniatures, Anneli Lax New Mathematical Library, Vol. 43, MAA. www.mathlinks.ro - Powered by www.gil.ro

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