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TOPICS IN INEQUALITIES Hojoo Lee Version 0.5 [2005/10/30] Introduction Inequalities are useful in all fields of Mathematics. The purpose in this book is to present standard techniques in the theory of inequalities. The readers will meet classical theorems including Schur’s inequality, Muirhead’s theorem, the Cauchy-Schwartz inequality, AM-GM inequality, and Hol̈der’s theorem, etc. There are many problems from Mathematical olympiads and competitions. The book is available at http://my.netian.com/∼ideahitme/eng.html I wish to express my appreciation to Stanley Rabinowitz who kindly sent me his paper On The Computer Solution of Symmetric Homogeneous Triangle Inequalities. This is an unfinished manuscript. I would greatly appreciate hearing about any errors in the book, even minor ones. You can send all comments to the author at [email protected] To Students The given techniques in this book are just the tip of the inequalities iceberg. What young students read this book should be aware of is that they should find their own creative methods to attack problems. It’s impossible to present all techniques in a small book. I don’t even claim that the methods in this book are mathematically beautiful. For instance, although Muirhead’s theorem and Schur’s theorem which can be found at chapter 3 are extremely powerful to attack homogeneous symmetric polynomial inequalities, it’s not a good idea for beginners to learn how to apply them to problems. (Why?) However, after mastering homogenization method using Muirhead’s theorem and Schur’s theorem, you can have a more broad mind in the theory of inequalities. That’s why I include the methods in this book. Have fun! Recommended Reading List 1. K. S. Kedlaya, A < B, http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html 2. I. Niven, Maxima and Minima Without Calculus, MAA 3. T. Andreescu, Z. Feng, 103 Trigonometry Problems From the Training of the USA IMO Team, Birkhauser 4. O. Bottema, R. Z̃. Djordjević, R. R. Janić, D. S. Mitrinović, P. M. Vasić, Geometric Inequalities, Wolters-Noordhoff Publishing, Groningen 1969 1 Contents 1 100 Problems 3 2 Substitutions 2.1 Euler’s Theorem and the Ravi Substitution 2.2 Trigonometric Substitutions . . . . . . . . 2.3 Algebraic Substitutions . . . . . . . . . . . 2.4 Supplementary Problems for Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 11 14 18 24 3 Homogenizations 3.1 Homogeneous Polynomial Inequalities . 3.2 Schur’s Theorem . . . . . . . . . . . . . 3.3 Muirhead’s Theorem . . . . . . . . . . . 3.4 Polynomial Inequalities with Degree 3 . 3.5 Supplementary Problems for Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 26 28 30 33 36 . . . . . Hölder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 37 39 43 44 . . . . . . . . . . 4 Normalizations 4.1 Normalizations . . . . . . . . . . . . . . . . . . . . 4.2 Classical Theorems : Cauchy-Schwartz, (Weighted) 4.3 Homogenizations and Normalizations . . . . . . . . 4.4 Supplementary Problems for Chapter 4 . . . . . . . . . . . . . . AM-GM, and . . . . . . . . . . . . . . . . 5 Multivariable Inequalities 45 6 References 53 2 Chapter 1 100 Problems Each problem that I solved became a rule, which served afterwards to solve other problems. Rene Descartes I 1. (Hungary 1996) (a + b = 1, a, b > 0) a2 b2 1 + ≥ a+1 b+1 3 I 2. (Columbia 2001) (x, y ∈ R) 3(x + y + 1)2 + 1 ≥ 3xy I 3. (0 < x, y < 1) xy + y x > 1 I 4. (APMC 1993) (a, b ≥ 0) Ã√ và u √ √ √ √ !3 √ !2 √ 3 3 3 2 2 2 + 3 b2 a+ b a + a b + ab + b a + ab + b u a ≤ ≤ ≤t 2 4 3 2 I 5. (Czech and Slovakia 2000) (a, b > 0) s r r µ ¶ 1 1 a 3 b 3 3 2(a + b) + ≥ + a b b a I 6. (Die √ W U RZEL, Heinz-Jürgen Seiffert) (xy > 0, x, y ∈ R) r 2xy x2 + y 2 x+y √ + ≥ xy + x+y 2 2 I 7. (Crux Mathematicorum, Problem 2645, Hojoo Lee) (a, b, c > 0) 2(a3 + b3 + c3 ) 9(a + b + c)2 + 2 ≥ 33 abc (a + b2 + c2 ) I 8. (x, y, z > 0) √ 3 I 9. (a, b, c, x, y, z > 0) xyz + x+y+z |x − y| + |y − z| + |z − x| ≥ 3 3 p 3 (a + x)(b + y)(c + z) ≥ 3 √ 3 abc + √ 3 xyz I 10. (x, y, z > 0) x+ x p (x + y)(x + z) + y+ y p (y + z)(y + x) + z+ p z (z + x)(z + y) ≤1 I 11. (x + y + z = 1, x, y, z > 0) ³ I 12. (Iran 1998) 1 x + 1 y + 1 z x y z √ +√ +√ ≥ 1−y 1−x 1−z ´ = 2, x, y, z > 1 √ √ x+y+z ≥ x−1+ r 3 2 p √ y−1+ z−1 I 13. (KMO Winter Program Test 2001) (a, b, c > 0) p p (a2 b + b2 c + c2 a) (ab2 + bc2 + ca2 ) ≥ abc + 3 (a3 + abc) (b3 + abc) (c3 + abc) I 14. (KMO Summer Program Test 2001) (a, b, c > 0) p p p p a4 + b4 + c4 + a2 b2 + b2 c2 + c2 a2 ≥ a3 b + b3 c + c3 a + ab3 + bc3 + ca3 I 15. (Gazeta Matematicã, Hojoo Lee) (a, b, c > 0) p p p p p p a4 + a2 b2 + b4 + b4 + b2 c2 + c4 + c4 + c2 a2 + a4 ≥ a 2a2 + bc + b 2b2 + ca + c 2c2 + ab I 16. (a, b, c ∈ R) p a2 I 17. (a, b, c > 0) + (1 − b)2 √ p p 3 2 2 2 2 2 + b + (1 − c) + c + (1 − a) ≥ 2 p a2 − ab + b2 + p b2 − bc + c2 ≥ p a2 + ac + c2 I 18. (Belarus 2002) (a, b, c, d > 0) p 2|ad − bc| (a + c)2 + (b + d)2 + p (a + c)2 + (b + d)2 ≥ p a2 + b2 + p c2 + d2 ≥ I 19. (Hong Kong 1998) (a, b, c ≥ 1) p √ √ √ a − 1 + b − 1 + c − 1 ≤ c(ab + 1) I 20. (Carlson’s inequality) (a, b, c > 0) r r ab + bc + ca 3 (a + b)(b + c)(c + a) ≥ 8 3 I 21. (Korea 1998) (x + y + z = xyz, x, y, z > 0) √ 1 1 1 3 +√ +p ≤ 2 2 2 2 1+x 1+z 1+y I 22. (IMO 2001) (a, b, c > 0) √ a b c +√ +√ ≥1 a2 + 8bc b2 + 8ca c2 + 8ab I 23. (IMO Short List 2004) (ab + bc + ca = 1, a, b, c > 0) r r r 1 3 1 3 1 3 1 + 6b + + 6c + + 6a ≤ a b c abc 4 p (a + c)2 + (b + d)2 I 24. (a, b, c > 0) p ab(a + b) + p bc(b + c) + I 25. (Macedonia 1995) (a, b, c > 0) r p a + b+c ca(c + a) ≥ r b + c+a p 4abc + (a + b)(b + c)(c + a) r c ≥2 a+b I 26. (Nesbitt’s inequality) (a, b, c > 0) a b c 3 + + ≥ b+c c+a a+b 2 I 27. (IMO 2000) (abc = 1, a, b, c > 0) µ ¶µ ¶µ ¶ 1 1 1 a−1+ b−1+ c−1+ ≤1 b c a I 28. ([ONI], Vasile Cirtoaje) (a, b, c > 0) µ ¶µ ¶ µ ¶µ ¶ µ ¶µ ¶ 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 I 29. (IMO Short List 1998) (xyz = 1, x, y, z > 0) x3 y3 z3 3 + + ≥ (1 + y)(1 + z) (1 + z)(1 + x) (1 + x)(1 + y) 4 I 30. (IMO Short List 1996) (abc = 1, a, b, c > 0) ab bc ca + + ≤1 a5 + b5 + ab b5 + c5 + bc c5 + a5 + ca I 31. (IMO 1995) (abc = 1, a, b, c > 0) 1 1 1 3 + 3 + 3 ≥ + c) b (c + a) c (a + b) 2 a3 (b I 32. (IMO Short List 1993) (a, b, c, d > 0) a b c d 2 + + + ≥ b + 2c + 3d c + 2d + 3a d + 2a + 3b a + 2b + 3c 3 I 33. (IMO Short List 1990) (ab + bc + cd + da = 1, a, b, c, d > 0) a3 b3 c3 d3 1 + + + ≥ b+c+d c+d+a d+a+b a+b+c 3 I 34. (IMO 1968) (x1 , x2 > 0, y1 , y2 , z1 , z2 ∈ R, x1 y1 > z1 2 , x2 y2 > z2 2 ) 1 1 8 + ≥ x1 y1 − z1 2 x2 y2 − z2 2 (x1 + x2 )(y1 + y2 ) − (z1 + z2 )2 I 35. (Romania 1997) (a, b, c > 0) a2 b2 c2 bc ca ab a2 + 2 + 2 ≥1≥ 2 + + + 2bc b + 2ca c + 2ab a + 2bc b2 + 2ca c2 + 2ab I 36. (Canada 2002) (a, b, c > 0) b3 c3 a3 + + ≥a+b+c bc ca ab 5 I 37. (USA 1997) (a, b, c > 0) 1 1 1 1 + + ≤ . a3 + b3 + abc b3 + c3 + abc c3 + a3 + abc abc I 38. (Japan 1997) (a, b, c > 0) (b + c − a)2 (c + a − b)2 (a + b − c)2 3 + + ≥ 2 2 2 2 (b + c) + a (c + a) + b (a + b)2 + c2 5 I 39. (USA 2003) (a, b, c > 0) (2a + b + c)2 (2b + c + a)2 (2c + a + b)2 + + ≤8 2a2 + (b + c)2 2b2 + (c + a)2 2c2 + (a + b)2 I 40. (Crux Mathematicorum, Problem 2580, Hojoo Lee) (a, b, c > 0) 1 1 1 b+c c+a a+b + + ≥ 2 + + a b c a + bc b2 + ca c2 + ab I 41. (Crux Mathematicorum, Problem 2581, Hojoo Lee) (a, b, c > 0) a2 + bc b2 + ca c2 + ab + + ≥a+b+c b+c c+a a+b I 42. (Crux Mathematicorum, Problem 2532, Hojoo Lee) (a2 + b2 + c2 = 1, a, b, c > 0) 1 1 2(a3 + b3 + c3 ) 1 + + ≥ 3 + a2 b2 c2 abc I 43. (Belarus 1999) (a2 + b2 + c2 = 3, a, b, c > 0) 1 1 1 3 + + ≥ 1 + ab 1 + bc 1 + ca 2 I 44. (Crux Mathematicorum, Problem 3032, Vasile Cirtoaje) (a2 + b2 + c2 = 1, a, b, c > 0) 1 1 1 9 + + ≤ 1 − ab 1 − bc 1 − ca 2 I 45. (Moldova 2005) (a4 + b4 + c4 = 3, a, b, c > 0) 1 1 1 + + ≤1 4 − ab 4 − bc 4 − ca I 46. (Greece 2002) (a2 + b2 + c2 = 1, a, b, c > 0) √ √ ´2 a b c 3³ √ + + ≥ a a + b b + c c b2 + 1 c2 + 1 a2 + 1 4 I 47. (Iran 1996) (a, b, c > 0) µ (ab + bc + ca) 1 1 1 + + (a + b)2 (b + c)2 (c + a)2 ¶ ≥ 9 4 I 48. (Albania 2002) (a, b, c > 0) √ µ ¶ p 1+ 3 2 1 1 1 √ (a + b2 + c2 ) + + ≥ a + b + c + a2 + b2 + c2 a b c 3 3 I 49. (Belarus 1997) (a, b, c > 0) c a+b b+c c+a a b + + ≥ + + b c a c+a a+b b+c 6 I 50. (Belarus 1998, I. Gorodnin) (a, b, c > 0) a b c a+b b+c + + ≥ + +1 b c a b+c a+b ¡ ¢ I 51. (Poland 1996) a + b + c = 1, a, b, c ≥ − 34 a b c 9 + + ≤ a2 + 1 b2 + 1 c2 + 1 10 I 52. (Bulgaria 1997) (abc = 1, a, b, c > 0) 1 1 1 1 1 1 + + ≤ + + 1+a+b 1+b+c 1+c+a 2+a 2+b 2+c I 53. (Romania 1997) (xyz = 1, x, y, z > 0) x9 + y 9 y9 + z9 z 9 + x9 + + ≥2 x6 + x3 y 3 + y 6 y6 + y3 z3 + z6 z 6 + z 3 x3 + x6 I 54. (Vietnam 1991) (x ≥ y ≥ z > 0) x2 y y 2 z z2x + + ≥ x2 + y 2 + z 2 z x y I 55. (Iran 1997) (x1 x2 x3 x4 = 1, x1 , x2 , x3 , x4 > 0) µ ¶ 1 1 1 1 x31 + x32 + x33 + x34 ≥ max x1 + x2 + x3 + x4 , + + + x1 x2 x3 x4 I 56. (Hong Kong 2000) (abc = 1, a, b, c > 0) 1 + ab2 1 + bc2 1 + ca2 18 + + ≥ 3 3 3 3 c a b a + b3 + c3 I 57. (Hong Kong 1997) (x, y, z > 0) p √ xyz(x + y + z + x2 + y 2 + z 2 ) 3+ 3 ≥ 9 (x2 + y 2 + z 2 )(xy + yz + zx) I 58. (Czech-Slovak Match 1999) (a, b, c > 0) a b c + + ≥1 b + 2c c + 2a a + 2b I 59. (Moldova 1999) (a, b, c > 0) ab bc ca a b c + + ≥ + + c(c + a) a(a + b) b(b + c) c+a b+a c+b I 60. (Baltic Way 1995) (a, b, c, d > 0) a+c b+d c+a d+b + + + ≥4 a+b b+c c+d d+a I 61. ([ONI], Vasile Cirtoaje) (a, b, c, d > 0) a−b b−c c−d d−a + + + ≥0 b+c c+d d+a a+b I 62. (Poland 1993) (x, y, u, v > 0) xy + xv + uy + uv xy uv ≥ + x+y+u+v x+y u+v 7 I 63. (Belarus 1997) (a, x, y, z > 0) a+y a+z a+x a+z a+x a+y x+ y+ z ≥x+y+z ≥ x+ y+ z a+x a+x a+y a+z a+y a+z I 64. (Lithuania 1987) (x, y, z > 0) x2 x3 y3 z3 x+y+z + 2 + 2 ≥ 2 2 + xy + y y + yz + z z + zx + x2 3 I 65. (Klamkin’s inequality) (−1 < x, y, z < 1) 1 1 + ≥2 (1 − x)(1 − y)(1 − z) (1 + x)(1 + y)(1 + z) I 66. (xy + yz + zx = 1, x, y, z > 0) x y z 2x(1 − x2 ) 2y(1 − y 2 ) 2z(1 − z 2 ) + + ≥ + + 1 + x2 1 + y2 1 + z2 (1 + x2 )2 (1 + y 2 )2 (1 + z 2 )2 I 67. (Russia 2002) (x + y + z = 3, x, y, z > 0) √ √ √ x + y + z ≥ xy + yz + zx I 68. (APMO 1998) (a, b, c > 0) ³ ¶ µ ¶ µ a´ b ³ c´ a+b+c 1+ 1+ 1+ ≥2 1+ √ 3 b c a abc I 69. (Elemente der Mathematik, Problem 1207, S̃efket Arslanagić) (x, y, z > 0) x y z x+y+z + + ≥ 3√ y z x xyz I 70. (Die √ W U RZEL, Walther Janous) (x + y + z = 1, x, y, z > 0) (1 + x)(1 + y)(1 + z) ≥ (1 − x2 )2 + (1 − y 2 )2 + (1 − z 2 )2 I 71. (United Kingdom 1999) (p + q + r = 1, p, q, r > 0) 7(pq + qr + rp) ≤ 2 + 9pqr I 72. (USA 1979) (x + y + z = 1, x, y, z > 0) x3 + y 3 + z 3 + 6xyz ≥ 1 . 4 I 73. (IMO 1984) (x + y + z = 1, x, y, z ≥ 0) 0 ≤ xy + yz + zx − 2xyz ≤ 7 27 I 74. (IMO Short List 1993) (a + b + c + d = 1, a, b, c, d > 0) abc + bcd + cda + dab ≤ 176 1 + abcd 27 27 I 75. (Poland 1992) (a, b, c ∈ R) (a + b − c)2 (b + c − a)2 (c + a − b)2 ≥ (a2 + b2 − c2 )(b2 + c2 − a2 )(c2 + a2 − b2 ) 8 I 76. (Canada 1999) (x + y + z = 1, x, y, z ≥ 0) x2 y + y 2 z + z 2 x ≤ 4 27 I 77. (Hong Kong 1994) (xy + yz + zx = 1, x, y, z > 0) √ 4 3 x(1 − y )(1 − z ) + y(1 − z )(1 − x ) + z(1 − x )(1 − y ) ≤ 9 2 2 2 2 2 2 I 78. (Vietnam 1996) (2(ab + ac + ad + bc + bd + cd) + abc + bcd + cda + dab = 16, a, b, c, d ≥ 0) a+b+c+d≥ 2 (ab + ac + ad + bc + bd + cd) 3 ¡ I 79. (Poland 1998) a + b + c + d + e + f = 1, ace + bdf ≥ 1 108 ¢ a, b, c, d, e, f > 0 abc + bcd + cde + def + ef a + f ab ≤ 1 36 I 80. (Italy 1993) (0 ≤ a, b, c ≤ 1) a2 + b2 + c2 ≤ a2 b + b2 c + c2 a + 1 I 81. (Czech Republic 2000) (m, n ∈ N, x ∈ [0, 1]) (1 − xn )m + (1 − (1 − x)m )n ≥ 1 I 82. (Ireland 1997) (a + b + c ≥ abc, a, b, c ≥ 0) a2 + b2 + c2 ≥ abc I 83. (BMO 2001) (a + b + c ≥ abc, a, b, c ≥ 0) a2 + b2 + c2 ≥ I 84. (Bearus 1996) (x + y + z = √ √ 3abc xyz, x, y, z > 0) xy + yz + zx ≥ 9(x + y + z) I 85. (Poland 1991) (x2 + y 2 + z 2 = 2, x, y, z ∈ R) x + y + z ≤ 2 + xyz I 86. (Mongolia 1991) (a2 + b2 + c2 = 2, a, b, c ∈ R) √ |a3 + b3 + c3 − abc| ≤ 2 2 I 87. (Vietnam 2002, Dung Tran Nam) (a2 + b2 + c2 = 9, a, b, c ∈ R) 2(a + b + c) − abc ≤ 10 I 88. (Vietnam 1996) (a, b, c > 0) (a + b)4 + (b + c)4 + (c + a)4 ≥ ¢ 4¡ 4 a + b4 + c4 7 I 89. (x, y, z ≥ 0) ³ I 90. (Latvia 2002) 1 1+a4 xyz ≥ (y + z − x)(z + x − y)(x + y − z) ´ 1 1 1 + 1+b 4 + 1+c4 + 1+d4 = 1, a, b, c, d > 0 abcd ≥ 3 9 I 91. (Proposed for 1999 USAMO, [AB, pp.25]) (x, y, z > 1) xx 2 +2yz y 2 +2zx z 2 +2xy y z ≥ (xyz)xy+yz+zx I 92. (APMO 2004) (a, b, c > 0) (a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca) I 93. (USA 2004) (a, b, c > 0) (a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a + b + c)3 I 94. (USA 2001) (a2 + b2 + c2 + abc = 4, a, b, c ≥ 0) 0 ≤ ab + bc + ca − abc ≤ 2 I 95. (Turkey, 1999) (c ≥ b ≥ a ≥ 0) (a + 3b)(b + 4c)(c + 2a) ≥ 60abc I 96. (Macedonia 1999) (a2 + b2 + c2 = 1, a, b, c > 0) a+b+c+ √ 1 ≥4 3 abc I 97. (Poland 1999) (a + b + c = 1, a, b, c > 0) √ a2 + b2 + c2 + 2 3abc ≤ 1 I 98. (Macedonia 2000) (x, y, z > 0) x2 + y 2 + z 2 ≥ √ 2 (xy + yz) I 99. (APMC 1995) (m, n ∈ N, x, y > 0) (n − 1)(m − 1)(xn+m + y n+m ) + (n + m − 1)(xn y m + xm y n ) ≥ nm(xn+m−1 y + xy n+m−1 ) I 100. ([ONI], Gabriel Dospinescu, Mircea Lascu, Marian Tetiva) (a, b, c > 0) a2 + b2 + c2 + 2abc + 3 ≥ (1 + a)(1 + b)(1 + c) 10 Chapter 2 Substitutions 2.1 Euler’s Theorem and the Ravi Substitution Many inequalities are simplified by some suitable substitutions. We begin with a classical inequality in triangle geometry. What is the first1 nontrivial geometric inequality ? In 1765, Euler showed that Theorem 1. Let R and r denote the radii of the circumcircle and incircle of the triangle ABC. Then, we have R ≥ 2r and the equality holds if and only if ABC is equilateral. Proof. Let BC = a, CA = b, AB = c, s = a+b+c and S = [ABC].2 Recall the well-known identities : 2 abc S S2 2 S = 4R , S = rs, S = s(s − a)(s − b)(s − c). Hence, R ≥ 2r is equivalent to abc 4S ≥ 2 s or abc ≥ 8 s or abc ≥ 8(s − a)(s − b)(s − c). We need to prove the following. Theorem 2. ([AP], A. Padoa) Let a, b, c be the lengths of a triangle. Then, we have abc ≥ 8(s − a)(s − b)(s − c) or abc ≥ (b + c − a)(c + a − b)(a + b − c) and the equality holds if and only if a = b = c. First Proof. We use the Ravi Substitution : Since a, b, c are the lengths of a triangle, there are positive reals x, y, z such that a = y + z, b = z + x, c = x + y. (Why?) Then, the inequality is (y + z)(z + x)(x + y) ≥ 8xyz for x, y, z > 0. However, we get (y + z)(z + x)(x + y) − 8xyz = x(y − z)2 + y(z − x)2 + z(x − y)2 ≥ 0. Second Proof. ([RI]) We may assume that a ≥ b ≥ c. It’s equivalent to a3 + b3 + c3 + 3abc ≥ a2 (b + c) + b2 (c + a) + c2 (a + b). Since c(a + b − c) ≥ b(c + a − b) ≥ c(a + b − c)3 , applying the Rearrangement inequality, we obtain a · a(b + c − a) + b · b(c + a − b) + c · c(a + b − c) ≤ a · a(b + c − a) + c · b(c + a − b) + a · c(a + b − c), a · a(b + c − a) + b · b(c + a − b) + c · c(a + b − c) ≤ c · a(b + c − a) + a · b(c + a − b) + b · c(a + b − c). Adding these two inequalities, we get the result. Exercise 1. Let ABC be a right triangle. Show that R ≥ (1 + √ 2)r. When does the equality hold ? It’s natural to ask that the inequality in the theorem 2 holds for arbitrary positive reals a, b, c? Yes ! It’s possible to prove the inequality without the additional condition that a, b, c are the lengths of a triangle : 1 The first geometric inequality is the Triangle Inequality : AB + BC ≥ AC this book, [P ] stands for the area of the polygon P . 3 For example, we have c(a + b − c) − b(c + a − b) = (b − c)(b + c − a) ≥ 0. 2 In 11 Theorem 3. Let x, y, z > 0. Then, we have xyz ≥ (y + z − x)(z + x − y)(x + y − z). The equality holds if and only if x = y = z. Proof. Since the inequality is symmetric in the variables, without loss of generality, we may assume that x ≥ y ≥ z. Then, we have x + y > z and z + x > y. If y + z > x, then x, y, z are the lengths of the sides of a triangle. And by the theorem 2, we get the result. Now, we may assume that y + z ≤ x. Then, xyz > 0 ≥ (y + z − x)(z + x − y)(x + y − z). The inequality in the theorem 2 holds when some of x, y, z are zeros : Theorem 4. Let x, y, z ≥ 0. Then, we have xyz ≥ (y + z − x)(z + x − y)(x + y − z). Proof. Since x, y, z ≥ 0, we can find positive sequences {xn }, {yn }, {zn } for which lim xn = x, lim yn = y, lim zn = z. n→∞ (For example, take xn = x + 1 n n→∞ n→∞ (n = 1, 2, · · · ), etc.) Applying the theorem 2 yields xn yn zn ≥ (yn + zn − xn )(zn + xn − yn )(xn + yn − zn ) Now, taking the limits to both sides, we get the result. Clearly, the equality holds when x = y = z. However, xyz = (y +z −x)(z +x−y)(x+y −z) and x, y, z ≥ 0 does not guarantee that x = y = z. In fact, for x, y, z ≥ 0, the equality xyz = (y + z − x)(z + x − y)(x + y − z) is equivalent to x = y = z or x = y, z = 0 or y = z, x = 0 or z = x, y = 0. It’s straightforward to verify the equality xyz − (y + z − x)(z + x − y)(x + y − z) = x(x − y)(x − z) + y(y − z)(y − x) + z(z − x)(z − y). Hence, the theorem 4 is a particular case of Schur’s inequality.4 Problem 1. (IMO 2000/2) Let a, b, c be positive numbers such that abc = 1. Prove that µ ¶µ ¶µ ¶ 1 1 1 a−1+ b−1+ c−1+ ≤ 1. b c a First Solution. Since abc = 1, we make the substitution a = xy , b = yz , c = xz for x, y, z > 0.5 We rewrite the given inequality in the terms of x, y, z : µ ¶ x z ³y x´ ³z y´ −1+ −1+ −1+ ≤ 1 ⇔ xyz ≥ (y + z − x)(z + x − y)(x + y − z). y y z z x x The Ravi Substitution is useful for inequalities for the lengths a, b, c of a triangle. After the Ravi Substitution, we can remove the condition that they are the lengths of the sides of a triangle. Problem 2. (IMO 1983/6) Let a, b, c be the lengths of the sides of a triangle. Prove that a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0. Solution. After setting a = y + z, b = z + x, c = x + y for x, y, z > 0, it becomes x3 z + y 3 x + z 3 y ≥ x2 yz + xy 2 z + xyz 2 or x2 y2 z2 + + ≥ x + y + z, y z x which follows from the Cauchy-Schwartz inequality µ 2 ¶ x y2 z2 (y + z + x) + + ≥ (x + y + z)2 . y z x 4 See 5 For the theorem 10 in the chapter 3. Take r = 1. 1 example, take x = 1, y = a1 , z = ab . 12 Problem 3. (IMO 1961/2, Weitzenböck’s inequality) Let a, b, c be the lengths of a triangle with area S. Show that √ a2 + b2 + c2 ≥ 4 3S. Solution. Write a = y + z, b = z + x, c = x + y for x, y, z > 0. It’s equivalent to ((y + z)2 + (z + x)2 + (x + y)2 )2 ≥ 48(x + y + z)xyz, which can be obtained as following : ((y + z)2 + (z + x)2 + (x + y)2 )2 ≥ 16(yz + zx + xy)2 ≥ 16 · 3(xy · yz + yz · zx + xy · yz).6 Exercise 2. (Hadwiger-Finsler inequality) Show that, for any triangle with sides a, b, c and area S, √ 2ab + 2bc + 2ca − (a2 + b2 + c2 ) ≥ 4 3S. Exercise 3. (Pedoe’s inequality) Let a1 , b1 , c1 denote the sides of the triangle A1 B1 C1 with area F1 . Let a2 , b2 , c2 denote the sides of the triangle A2 B2 C2 with area F2 . Show that a1 2 (a2 2 + b2 2 − c2 2 ) + b1 2 (b2 2 + c2 2 − a2 2 ) + c1 2 (c2 2 + a2 2 − b2 2 ) ≥ 16F1 F2 . 6 Here, we used the well-known inequalities p2 + q 2 ≥ 2pq and (p + q + r)2 ≥ 3(pq + qr + rp). 13 2.2 Trigonometric Substitutions If you are faced with an integral that contains square root expressions such as Z p Z p Z p 1 − x2 dx, 1 + y 2 dy, z 2 − 1 dz then trigonometric substitutions such as x = sin t, y = tan t, z = sec t are very useful. When dealing with square root expressions, making a suitable trigonometric substitution simplifies the given inequality. Problem 4. (Latvia 2002) Let a, b, c, d be the positive real numbers such that 1 1 1 1 + + + = 1. 4 4 4 1+a 1+b 1+c 1 + d4 Prove that abcd ≥ 3. ¢ ¡ Solution. We can write a2 = tan A, b2 = tan B, c2 = tan C, d2 = tan D, where A, B, C, D ∈ 0, π2 . Then, the algebraic identity becomes the following trigonometric identity : cos2 A + cos2 B + cos2 C + cos2 D = 1. Applying the AM-GM inequality, we obtain 2 sin2 A = 1 − cos2 A = cos2 B + cos2 C + cos2 D ≥ 3 (cos B cos C cos D) 3 . Similarly, we obtain 2 2 2 sin2 B ≥ 3 (cos C cos D cos A) 3 , sin2 C ≥ 3 (cos D cos A cos B) 3 , and sin2 D ≥ 3 (cos A cos B cos C) 3 . Multiplying these inequalities, we get the result! Exercise 4. ([ONI], Titu Andreescu, Gabriel Dosinescu) Let a, b, c, d be the real numbers such that (1 + a2 )(1 + b2 )(1 + c2 )(1 + d2 ) = 16. Prove that −3 ≤ ab + ac + ad + bc + bd + cd − abcd ≤ 5. Problem 5. (Korea 1998) Let x, y, z be the positive reals with x + y + z = xyz. Show that √ 1 1 1 3 +√ +p ≤ . 2 2 2 2 1+x 1+z 1+y Since the function f is not concave down on R+ , we cannot apply ¡ π ¢ Jensen’s inequality to the function 1 f (t) = √1+t . However, the function f (tan θ) is concave down on 0, 2 ! 2 ¡ ¢ Solution. We can write x = tan A, y = tan B, z = tan C, where A, B, C ∈ 0, π2 . Using the fact that ¢2 ¡ 1 + tan2 θ = cos1 θ , where cos θ 6= 0, we rewrite it in the terms of A, B, C : cos A + cos B + cos C ≤ 3 . 2 x+y It follows from tan(π − C) = −z = 1−xy = tan(A + B) and from π − C, A + B ∈ (0, π) that π − C = A + B or A + B + C = π. Hence, it suffices to show the following. Theorem 5. In any acute triangle ABC, we have cos A + cos B + cos C ≤ 23 . ¢ ¡ Proof. Since cos x is concave down on 0, π2 , it’s a direct consequence of Jensen’s inequality. ¢ ¡ We note that the function cos x is not concave down on (0, π). In fact, it’s concave up on π2 , π . One may think that the inequality cos A + cos B + cos C ≤ 32 doesn’t hold for any triangles. However, it’s known that it also holds for any triangles. 14 Theorem 6. In any triangle ABC, we have cos A + cos B + cos C ≤ 23 . First Proof. It follows from π − C = A + B that cos C = − cos(A + B) = − cos A cos B + sin A sin B or 3 − 2(cos A + cos B + cos C) = (sin A − sin B)2 + (cos A + cos B − 1)2 ≥ 0. Second Proof. Let BC = a, CA = b, AB = c. Use the Cosine Law to rewrite the given inequality in the terms of a, b, c : b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 3 + + ≤ . 2bc 2ca 2ab 2 Clearing denominators, this becomes 3abc ≥ a(b2 + c2 − a2 ) + b(c2 + a2 − b2 ) + c(a2 + b2 − c2 ), which is equivalent to abc ≥ (b + c − a)(c + a − b)(a + b − c) in the theorem 2. In case even when there is no condition such as x + y + z = xyz or xy + yz + zx = 1, the trigonometric substitutions are useful. Problem 6. (APMO 2004/5) Prove that, for all positive real numbers a, b, c, (a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca). √ √ √ ¡ ¢ Proof. Choose A, B, C ∈ 0, π2 with a = 2 tan A, b = 2 tan B, and c = 2 tan C. Using the well-known trigonometric identity 1 + tan2 θ = cos12 θ , one may rewrite it as 4 ≥ cos A cos B cos C (cos A sin B sin C + sin A cos B sin C + sin A sin B cos C) . 9 One may easily check the following trigonometric identity cos(A + B + C) = cos A cos B cos C − cos A sin B sin C − sin A cos B sin C − sin A sin B cos C. Then, the above trigonometric inequality takes the form 4 ≥ cos A cos B cos C (cos A cos B cos C − cos(A + B + C)) . 9 Let θ = A+B+C . 3 Applying the AM-GM inequality and Jesen’s inequality, we have µ cos A cos B cos C ≤ We now need to show that cos A + cos B + cos C 3 ¶3 ≤ cos3 θ. 4 ≥ cos3 θ(cos3 θ − cos 3θ). 9 Using the trigonometric identity cos 3θ = 4 cos3 θ − 3 cos θ or cos 3θ − cos 3θ = 3 cos θ − 3 cos3 θ, it becomes ¡ ¢ 4 ≥ cos4 θ 1 − cos2 θ , 27 which follows from the AM-GM inequality µ ¶1 ¶ µ ¢ 3 ¢ cos2 θ cos2 θ ¡ 1 1 cos2 θ cos2 θ ¡ 2 2 = . · · 1 − cos θ + + 1 − cos θ ≤ 2 2 3 2 2 3 One find that the equality holds if and only if tan A = tan B = tan C = 15 √1 2 if and only if a = b = c = 1. Exercise 5. ([TZ], pp.127) Let x, y, z be real numbers such that 0 < x, y, z < 1 and xy + yz + zx = 1. Prove that √ x y z 3 3 + + ≥ . 1 − x2 1 − y2 1 − z2 2 Exercise 6. ([TZ], pp.127) Let x, y, z be positive real numbers such that x + y + z = xyz. Prove that √ x y z 3 3 √ +p +√ ≤ . 2 1 + x2 1 + z2 1 + y2 Exercise 7. ([ONI], Florina Carlan, Marian Tetiva) Prove that if x, y, z > 0 satisfy the condition x + y + z = xyz then p p p xy + yz + zx ≥ 3 + 1 + x2 + 1 + y 2 + 1 + z 2 . Exercise 8. ([ONI], Gabriel Dospinescu, Marian Tetiva) Let x, y, z be positive real numbers such that x + y + z = xyz. Prove that √ (x − 1)(y − 1)(z − 1) ≤ 6 3 − 10. Exercise 9. ([TZ], pp.113) Let a, b, c be real numbers. Prove that (a2 + 1)(b2 + 1)(c2 + 1) ≥ (ab + bc + ca − 1)2 . Exercise 10. ([TZ], pp.149) Let a and b be positive real numbers. Prove that √ 1 2 1 +√ ≥√ 2 2 1 + ab 1+a 1+b if either (1) 0 < a, b ≤ 1 or (2) ab ≥ 3. In the theorem 1 and 2, we see that the geometric inequality R ≥ 2r is equivalent to the algebraic inequality abc ≥ (b + c − a)(c + a − b)(a + b − c). We now find that, in the proof of the theorem 6, abc ≥ (b + c − a)(c + a − b)(a + b − c) is equivalent to the trigonometric inequality cos A + cos B + cos C ≤ 32 . One may ask that In any triangles ABC, is there a natural relation between cos A + cos B + cos C and and r are the radii of the circumcircle and incircle of ABC ? R r, where R Theorem 7. Let R and r denote the radii of the circumcircle and incircle of the triangle ABC. Then, we have cos A + cos B + cos C = 1 + Rr . Proof. Use the identity a(b2 + c2 − a2 ) + b(c2 + a2 − b2 ) + c(a2 + b2 − c2 ) = 2abc + (b + c − a)(c + a − b)(a + b − c). We leave the details for the readers. Exercise 11. Let R and r be the radii of the circumcircle and incircle of the triangle ABC with BC = a, CA = b, AB = c. Let s denote the semiperimeter of ABC. Verify the follwing identities 7 : (1) ab + bc + ca = s2 + 4Rr + r2 , (2) abc = 4Rrs, (3) cos A cos B + cos B cos C + cos C cos A = (4) cos A cos B cos C = 2 2 s −(2R+r) 4R2 s2 −4R2 +r 2 , 4R2 Exercise 12. (a) Let p, q, r be the positive real numbers such that p2 + q 2 + r2 + 2pqr = 1. Show that there exists an acute triangle ABC such that p = cos A, q = cos B, r = cos C. ¤ £ (b) Let p, q, r ≥ 0 with p2 + q 2 + r2 + 2pqr = 1. Show that there are A, B, C ∈ 0, π2 with p = cos A, q = cos B, r = cos C, and A + B + C = π. 7 For more identities, see the exercise 10. 16 Exercise 13. ([ONI], Marian Tetiva) Let x, y, z be positive real numbers satisfying the condition x2 + y 2 + z 2 + 2xyz = 1. Prove that (1) (2) (3) (4) xyz ≤ 18 , xy + yz + zx ≤ 34 , x2 + y 2 + z 2 ≥ 34 , and xy + yz + zx ≤ 2xyz + 21 . Exercise 14. ([ONI], Marian Tetiva) Let x, y, z be positive real numbers satisfying the condition x2 + y 2 + z 2 = xyz. Prove that (1) (2) (3) (4) xyz ≥ 27, xy + yz + zx ≥ 27, x + y + z ≥ 9, and xy + yz + zx ≥ 2(x + y + z) + 9. Problem 7. (USA 2001) Let a, b, and c be nonnegative real numbers such that a2 + b2 + c2 + abc = 4. Prove that 0 ≤ ab + bc + ca − abc ≤ 2. Solution. Notice that a, b, c > 1 implies that a2 + b2 + c2 + abc > 4. If a ≤ 1, then we have ab + bc + ca − abc ≥ (1 − a)bc ≥ 0. We now prove that ab + bc + ca − abc ≤ 2. Letting a = 2p, b = 2q, c = 2r, we get p2 + q 2 + r2 + 2pqr = 1. By the exercise 12, we can write h πi a = 2 cos A, b = 2 cos B, c = 2 cos C for some A, B, C ∈ 0, with A + B + C = π. 2 We are required to prove cos A cos B + cos B cos C + cos C cos A − 2 cos A cos B cos C ≤ One may assume that A ≥ π 3 1 . 2 or 1 − 2 cos A ≥ 0. Note that cos A cos B + cos B cos C + cos C cos A − 2 cos A cos B cos C = cos A(cos B + cos C) + cos B cos C(1 − 2 cos A). We apply Jensen’s inequality to deduce cos B + cos C ≤ cos(B + C) ≤ 1 − cos A. These imply that 3 2 − cos A. Note that 2 cos B cos C = cos(B − C) + µ cos A(cos B + cos C) + cos B cos C(1 − 2 cos A) ≤ cos A However, it’s easy to verify that cos A ¡3 2 ¶ µ ¶ 3 1 − cos A − cos A + (1 − 2 cos A). 2 2 ¢ ¡ ¢ A − cos A + 1−cos (1 − 2 cos A) = 21 . 2 In the above solution, we showed that cos A cos B + cos B cos C + cos C cos A − 2 cos A cos B cos C ≤ 1 2 holds for all acute triangles. Using the results (c) and (d) in the exercise (4), we can rewrite it in the terms of R, r, s : 2R2 + 8Rr + 3r2 ≤ s2 . In 1965, W. J. Blundon found the best possible inequalities of the form A(R, r) ≤ s2 ≤ B(R, r), where A(x, y) and B(x, y) are real quadratic forms αx2 + βxy + γy 2 : 8 Exercise 15. Let R and r denote the radii of the circumcircle and incircle of the triangle ABC. Let s be the semiperimeter of ABC. Show that 16Rr − 5r2 ≤ s2 ≤ 4R2 + 4Rr + 3r2 . 8 For a proof, see [WJB]. 17 2.3 Algebraic Substitutions We know that some inequalities in triangle geometry can be treated by the Ravi substitution and trigonometric substitutions. We can also transform the given inequalities into easier ones through some clever algebraic substitutions. Problem 8. (IMO 2001/2) Let a, b, c be positive real numbers. Prove that √ a b c +√ +√ ≥ 1. a2 + 8bc b2 + 8ca c2 + 8ab First Solution. To remove the square roots, we make the following substitution : x= √ a2 a b c , y=√ , z=√ . 2 2 + 8bc b + 8ca c + 8ab Clearly, x, y, z ∈ (0, 1). Our aim is to show that x + y + z ≥ 1. We notice that µ ¶µ ¶µ 2 ¶ x2 y2 z2 b2 c2 1 x2 y2 z a2 = , = , = =⇒ = . 8bc 1 − x2 8ac 1 − y 2 8ab 1 − z2 512 1 − x2 1 − y2 1 − z2 Hence, we need to show that x + y + z ≥ 1, where 0 < x, y, z < 1 and (1 − x2 )(1 − y 2 )(1 − z 2 ) = 512(xyz)2 . However, 1 > x + y + z implies that, by the AM-GM inequality, (1 − x2 )(1 − y 2 )(1 − z 2 ) > ((x + y + z)2 − x2 )((x + y + z)2 − y 2 )((x + y + z)2 − z 2 ) = (x + x + y + z)(y + z) 1 1 1 1 1 1 (x + y + y + z)(z + x)(x + y + z + z)(x + y) ≥ 4(x2 yz) 4 · 2(yz) 2 · 4(y 2 zx) 4 · 2(zx) 2 · 4(z 2 xy) 4 · 2(xy) 2 = 512(xyz)2 . This is a contradiction ! Problem 9. (IMO 1995/2) Let a, b, c be positive numbers such that abc = 1. Prove that 1 1 1 3 + 3 + 3 ≥ . + c) b (c + a) c (a + b) 2 a3 (b First Solution. After the substitution a = x1 , b = y1 , c = z1 , we get xyz = 1. The inequality takes the form x2 y2 z2 3 + + ≥ . y+z z+x x+y 2 It follows from the Cauchy-schwartz inequality that µ 2 ¶ y2 z2 x [(y + z) + (z + x) + (x + y)] + + ≥ (x + y + z)2 y+z z+x x+y so that, by the AM-GM inequality, 1 x2 y2 z2 x+y+z 3(xyz) 3 3 + + ≥ ≥ = . y+z z+x x+y 2 2 2 We offer an alternative solution of the problem 5 : (Korea 1998) Let x, y, z be the positive reals with x + y + z = xyz. Show that √ 1 1 1 3 +√ +p ≤ . 2 2 2 2 1+x 1+z 1+y 18 Second Solution. The starting point is letting a = x1 , b = y1 , c = z1 . We find that a + b + c = abc is equivalent to 1 = xy + yz + zx. The inequality becomes √ or p y 3 x z +p ≤ +√ 2 2 2 x2 + 1 z + 1 y +1 x x2 + xy + yz + zx or y2 x p y +p (x + y)(x + z) + xy + yz + zx y +p (y + z)(y + x) +p z z2 + xy + yz + zx z +p (z + x)(z + y) ≤ ≤ 3 2 3 . 2 By the AM-GM inequality, we have p µ ¶ x (x + y)(x + z) 1 x[(x + y) + (x + z)] 1 x x x p ≤ = + . = (x + y)(x + z) 2 (x + y)(x + z) 2 x+z x+z (x + y)(x + z) In a like manner, we obtain y 1 p ≤ 2 (y + z)(y + x) µ y y + y+z y+x ¶ z 1 and p ≤ 2 (z + x)(z + y) µ z z + z+x z+y Adding these three yields the required result. We now prove a classical theorem in various ways. Theorem 8. (Nesbitt, 1903) For all positive real numbers a, b, c, we have a b c 3 + + ≥ . b+c c+a a+b 2 Proof 1. After the substitution x = b + c, y = c + a, z = a + b, it becomes X y+z−x X y+z 3 ≥ or ≥ 6, 2x 2 x cyclic cyclic which follows from the AM-GM inequality as following: µ ¶ X y+z y z z x x y y z z x x y 6 = + + + + + ≥6 · · · · · = 6. x x x y y z z x x y y z z 1 cyclic Proof 2. We make the substitution x= It follows that X f (x) = cyclic a b c , y= , z= . b+c c+a a+b X cyclic a t = 1, where f (t) = . a+b+c 1+t Since f is concave down on (0, ∞), Jensen’s inequality shows that µ ¶ µ ¶ µ ¶ µ ¶ 1 2 1 X x+y+z 1 x+y+z f = = f (x) ≥ f or f ≥f . 2 3 3 3 2 3 cyclic Since f is monotone decreasing, we have X a 1 x+y+z 3 ≤ or =x+y+z ≥ . 2 3 b+c 2 cyclic 19 ¶ . Proof 3. As in the previous proof, it suffices to show that T ≥ X x 1 x+y+z where T = and = 1. 2 3 1+x cyclic One can easily check that the condition X cyclic x =1 1+x becomes 1 = 2xyz + xy + yz + zx. By the AM-GM inequality, we have 1 = 2xyz + xy + yz + zx ≤ 2T 3 + 3T 2 ⇔ 2T 3 + 3T 2 − 1 ≥ 0 ⇔ (2T − 1)(T + 1)2 ≥ 0 ⇔ T ≥ 1 . 2 Proof 4. Since the inequality is symmetric in the three variables, we may assume that a ≥ b ≥ c. After the substitution x = ac , y = cb , we have x ≥ y ≥ 1. It becomes a c b c +1 + a c b c +1 + a c 1 + b c ≥ 3 x y 3 1 or + ≥ − . 2 y+1 x+1 2 x+y We apply the AM-GM inequality to obtain x+1 y+1 x y 1 1 + ≥ 2 or + ≥2− + . y+1 x+1 y+1 x+1 y+1 x+1 It suffices to show that 2− 1 1 3 1 1 1 1 1 y−1 y−1 + ≥ − ⇔ − ≥ − ⇔ ≥ . y+1 x+1 2 x+y 2 y+1 x+1 x+y 2(1 + y) (x + 1)(x + y) However, the last inequality clearly holds for x ≥ y ≥ 1. Proof 5. As in the previous proof, we need to prove x y 1 3 + + ≥ where x ≥ y ≥ 1. y+1 x+1 x+y 2 Let A = x + y and B = xy. It becomes x2 + y 2 + x + y 1 3 A2 − 2B + A 1 3 + ≥ or + ≥ or 2A3 − A2 − A + 2 ≥ B(7A − 2). (x + 1)(y + 1) x+y 2 A+B+1 A 2 Since 7A − 2 > 2(x + y − 1) > 0 and A2 = (x + y)2 ≥ 4xy = 4B, it’s enough to show that 4(2A3 − A2 − A + 2) ≥ A2 (7A − 2) ⇔ A3 − 2A2 − 4A + 8 ≥ 0. However, it’s easy to check that A3 − 2A2 − 4A + 8 = (A − 2)2 (A + 2) ≥ 0. We now present alternative solutions of problem 1. (IMO 2000/2) Let a, b, c be positive numbers such that abc = 1. Prove that µ ¶µ ¶µ ¶ 1 1 1 a−1+ b−1+ c−1+ ≤ 1. b c a Second Solution. ([IV], Ilan Vardi) Since abc = 1, we may assume that a ≥ 1 ≥ b. 9 It follows that µ ¶µ ¶µ ¶ µ ¶µ ¶ 1 1 1 1 1 (a − 1)(1 − b) 10 1− a−1+ b−1+ c−1+ = c+ −2 a+ −1 + . b c a c b a 9 Why? 10 For Note that the inequality is not symmetric in the three variables. Check it! a verification of the identity, see [IV]. 20 Third Solution. As in the first solution, after the substitution a = xy , b = yz , c = xz for x, y, z > 0, we can rewrite it as xyz ≥ (y + z − x)(z + x − y)(x + y − z). Without loss of generality, we can assume that z ≥ y ≥ x. Set y − x = p and z − x = q with p, q ≥ 0. It’s straightforward to verify that xyz ≥ (y + z − x)(z + x − y)(x + y − z) = (p2 − pq + q 2 )x + (p3 + q 3 − p2 q − pq 2 ). Since p2 − pq + q 2 ≥ (p − q)2 ≥ 0 and p3 + q 3 − p2 q − pq 2 = (p − q)2 (p + q) ≥ 0, we get the result. Fourth Solution. (based on work by an IMO 2000 contestant from Japan) Putting c = µ ¶ µ ¶ 1 1 1 a−1+ (b − 1 + ab) −1+ ≤1 b ab a or 1 ab , it becomes a3 b3 − a2 b3 − ab3 − a2 b2 + 3ab2 − ab + b3 − b2 − b + 1 ≥ 0. Setting x = ab, it becomes f (x) ≥ 0, where fb (t) = t3 + b3 − b2 t − bt2 + 3bt − t2 − b2 − t − b + 1. Fix a positive number b ≥ 1. We need to show that F (t) := fb (t) ≥ 0 for all t ≥ 0. It’s easy to check that the cubic polynomial F / (t) = 3t2 − 2(b + 1)t − (b2 − 3b + 1) has two real roots √ √ b + 1 + 4b2 − 7b + 4 b + 1 − 4b2 − 7b + 4 and λ = . 3 3 Since F has a local minimum at t = λ, we find that F (t) ≥ M in {F (0), F (λ)} for all t ≥ 0. We have to prove that F (0) ≥ 0 and F (λ) ≥ 0. Since F (0) = b3 − b2 − b + 1 = (b − 1)2 (b + 1) ≥ 0, it remains to show that F (λ) ≥ 0. Notice that λ is a root of F / (t). After long division, we get µ ¶ ¢ 1 b+1 1¡ F (t) = F / (t) t− + (−8b2 + 14b − 8)t + 8b3 − 7b2 − 7b + 8 . 3 9 9 Putting t = λ, we have ¢ 1¡ (−8b2 + 14b − 8)λ + 8b3 − 7b2 − 7b + 8 . 9 Thus, our job is now to establish that, for all b ≥ 0, ! à √ b + 1 + 4b2 − 7b + 4 2 (−8b + 14b − 8) + 8b3 − 7b2 − 7b + 8 ≥ 0, 3 F (λ) = which is equivalent to p 16b3 − 15b2 − 15b + 16 ≥ (8b2 − 14b + 8) 4b2 − 7b + 4 . Since both 16b3 − 15b2 − 15b + 16 and 8b2 − 14b + 8 are positive,11 it’s equivalent to (16b3 − 15b2 − 15b + 16)2 ≥ (8b2 − 14b + 8)2 (4b2 − 7b + 4) or 864b5 − 3375b4 + 5022b3 − 3375b2 + 864b ≥ 0 or 864b4 − 3375b3 + 5022b2 − 3375b + 864 ≥ 0. Let G(x) = 864×4 − 3375×3 + 5022×2 − 3375x + 864. We prove that G(x) ≥ 0 for all x ∈ R. We find that G/ (x) = 3456×3 − 10125×2 + 10044x − 3375 = (x − 1)(3456×2 − 6669x + 3375). Since 3456×2 − 6669x + 3375 > 0 for all x ∈ R, we find that G(x) and x − 1 have the same sign. It follows that G(x) is monotone decreasing on (−∞, 1] and monotone increasing on [1, ∞). We conclude that G(x) has the global minimum at x = 1. Hence, G(x) ≥ G(1) = 0 for all x ∈ R. 11 It’s easy to check that 16b3 − 15b2 − 15b + 16 = 16(b3 − b2 − b + 1) + b2 + b > 16(b2 − 1)(b − 1) ≥ 0 and 8b2 − 14b + 8 = 8(b − 1)2 + 2b > 0. 21 Fifth Solution. (From the IMO 2000 Short List) Using the condition abc = 1, it’s straightforward to verify the equalities µ ¶ µ ¶ 1 1 1 2= a−1+ +c b−1+ , a b c µ ¶ µ ¶ 1 1 1 2= b−1+ +a c−1+ , b c a µ ¶ µ ¶ 1 1 1 2= c−1+ +b a−1+ . c a c In particular, they show that at most one of the numbers u = a − 1 + 1b , v = b − 1 + 1c , w = c − 1 + negative. If there is such a number, we have µ ¶µ ¶µ ¶ 1 1 1 a−1+ b−1+ c−1+ = uvw < 0 < 1. b c a 1 a is And if u, v, w ≥ 0, the AM-GM inequality yields r r r 1 c 1 a 1 b uv, 2 = v + aw ≥ 2 vw, 2 = w + aw ≥ 2 wu. 2 = u + cv ≥ 2 a a b b c c Thus, uv ≤ ac , vw ≤ ab , wu ≤ cb , so (uvw)2 ≤ a c · ab · cb = 1. Since u, v, w ≥ 0, this completes the proof. It turns out that the substitution p = x + y + z, q = xy + yz + zx, r = xyz is powerful for the three variables inequalities. We need the following lemma. Lemma 1. Let x, y, z be non-negative real numbers numbers. Set p = x + y + z, q = xy + yz + zx, and r = xyz. Then, we have 12 (1) p3 − 4pq + 9r ≥ 0, (2) p4 − 5p2 q + 4q 2 + 6pr ≥ 0, (3) pq − 9r ≥ 0. Proof. They are equivalent to (10 ) x(x − y)(x − z) + y(y − z)(y − x) + z(z − x)(z − y) ≥ 0, (20 ) x2 (x − y)(x − z) + y 2 (y − z)(y − x) + z 2 (z − x)(z − y) ≥ 0,13 (30 ) x(y − z)2 + y(z − x)2 + z(x − y)2 ≥ 0. We leave the details for the readers. Problem 10. (Iran 1996) Let x, y, z be positive real numbers. Prove that µ ¶ 1 1 1 9 (xy + yz + zx) + + ≥ . 2 2 2 (x + y) (y + z) (z + x) 4 First Solution. We make the substitution p = x + y + z, q = xy + yz + zx, r = xyz. Notice that (x + y)(y + z)(z + x) = (x + y + z)(xy + yz + zx) − xyz = pq − r. One may easily rewrite the given inequality in the terms of p, q, r : ¶ µ 2 9 (p + q)2 − 4p(pq − r) ≥ q (pq − r)2 4 or 4p4 q − 17p2 q 2 + 4q 3 + 34pqr − 9r2 ≥ 0 or pq(p3 − 4pq + 9r) + q(p4 − 5p2 q + 4q 2 + 6pr) + r(pq − 9r) ≥ 0. We find that every term on the left hand side is nonnegative by the lemma. 12 When 13 See does equality hold in each inequality? For more p-q-r inequalities, visit the site [ESF]. the theorem 10. 22 Problem 11. Let x, y, z be nonnegative real numbers with xy + yz + zx = 1. Prove that 1 1 1 5 + + ≥ . x+y y+z z+x 2 First Solution. Rewrite the inequality in the terms of p = x + y + z, q = xy + yz + zx, r = xyz: 4p4 q + 4q 3 − 17p2 q 2 − 25r2 + 50pqr ≥ 0. It can be rewritten as 3pq(p3 − 4pq + 9r) + q(p4 − 5p2 q + 4q 2 + 6pr) + 17r(pq − 9r) + 128r2 ≥ 0. However, the every term on the left hand side is nonnegative by the lemma. Exercise 16. (Carlson’s inequality) Prove that, for all positive real numbers a, b, c, r r ab + bc + ca 3 (a + b)(b + c)(c + a) ≥ . 8 3 Exercise 17. (Bulgaria 1997) Let a, b, c be positive real numbers such that abc = 1. Prove that 1 1 1 1 1 1 + + ≤ + + . 1+a+b 1+b+c 1+c+a 2+a 2+b 2+c We close this section by presenting a problem which can be solved by two algebraic substitutions and a trigonometric substitution. Problem 12. (Iran 1998) Prove that, for all x, y, z > 1 such that + 1 y + 1 z = 2, p √ y − 1 + z − 1. √ √ √ First Solution. We begin with the algebraic substitution a = x − 1, b = y − 1, c = z − 1. Then, the condition becomes √ x+y+z ≥ √ 1 x x−1+ 1 1 1 + + = 2 ⇔ a2 b2 + b2 c2 + c2 a2 + 2a2 b2 c2 = 1 2 2 1+a 1+b 1 + c2 and the inequality is equivalent to p 3 a2 + b2 + c2 + 3 ≥ a + b + c ⇔ ab + bc + ca ≤ . 2 Let p = bc, q = ca, r = ab. Our job is to prove that p + q + r ≤ exercise 12, we can make the trigonometric substitution 3 2 where p2 + q 2 + r2 + 2pqr = 1. By the h π´ p = cos A, q = cos B, r = cos C for some A, B, C ∈ 0, with A + B + C = π. 2 What we need to show is now that cos A+cos B+cos C ≤ 32 . However, it follows from Jensen’s inequality! 23 2.4 Supplementary Problems for Chapter 2 Exercise 18. Let x, y, and z be positive numbers. Let p = x + y + z, q = xy + yz + zx, and r = xyz. Prove the following inequalities : (a) p2 ≥ 3q (b) p3 ≥ 27r (c) q 2 ≥ 3pr (d) 2p3 + 9r ≥ 7pq (e) p2 q + 3pr ≥ 4q 2 (f ) p2 q ≥ 3pr + 2q 2 (g) p4 + 3q 2 ≥ 4p2 q (h) pq 2 ≥ 2p2 r + 3qr (i) 2q 3 + 9r3 ≥ 7pqr (j) q 3 + 9r2 ≥ 4pqr (k) p3 r + q 3 ≥ 6pqr Exercise 19. ([ONI], Mircea Lascu, Marian Tetiva) Let x, y, z be positive real numbers satisfying the condition xy + yz + zx + 2xyz = 1. Prove that (1) xyz ≤ 81 , (2) x + y + z ≤ 32 , (3) x1 + y1 + z1 ≥ 4(x + y + z), and (4) 1 x + 1 y + 1 z − 4(x + y + z) ≥ (2z−1)2 z(2z+1) , where z ≥ x, y. Exercise 20. Let f (x, y) be a real polynomial such that, for all θ ∈ R3 , f (cos θ, sin θ) = 0. Show that the polynomial f (x, y) is divisible by x2 + y 2 − 1. Exercise 21. Let f (x, y, z) be a real polynomial. Suppose that f (cos α, cos β, cos γ) = 0, for all α, β, γ ∈ R3 with α + β + γ = π. Show that f (x, y, z) is divisible by x2 + y 2 + z 2 + 2xyz − 1. 14 Exercise 22. (IMO Unused 1986) Let a, b, c be positive real numbers. Show that (a + b − c)2 (a − b + c)2 (−a + b + c)2 ≥ (a2 + b2 − c2 )(a2 − b2 + c2 )(−a2 + b2 + c2 ). 15 Exercise 23. With the usual notation for a triangle, verify the following identities : (1) sin A + sin B + sin C = Rs (2) sin A sin B + sin B sin C + sin C sin A = sr (3) sin A sin B sin C = 2R 2 s2 +4Rr+r 2 4R2 s(s2 −6Rr−3r 2 ) 4R3 3 −3rs2 −4R3 = (2R+r) 4R 3 (4) sin3 A + sin3 B + sin3 C = (5) cos3 A + cos3 B + cos3 C (6) tan A + tan B + tan C = tan A tan B tan C = (7) tan A tan B + tan B tan C + tan C tan A = (8) cot A + cot B + cot C = r (9) sin A2 sin B2 sin C2 = 4R s (10) cos A2 cos B2 cos C2 = 4R 2rs s2 −(2R+r)2 s −4Rr−r 2 s2 −(2R+r)2 2 s2 −4Rr−r 2 2sr 14 For a proof, see [JmhMh]. we assume that there is a triangle ABC with BC = a, CA = b, AB = c, then it’s equivalent to the inequality s2 ≤ 4R2 + 4Rr + 3r2 in the exercise 6. 15 If 24 Exercise 24. Let a, b, c be the lengths of the sides of a triangle. Let s be the semi-perimeter of the triangle. Then, the following inequalities holds. 2 (a) 3(ab + bc + ca) ≤ (a + b + c) ¡ 2 < 4(ab ¢ + bc + ca) 36 2 2 2 (b) [JfdWm] a + b + c ≥ 35 s + abc s (c) [AP] 8(s − a)(s − b)(s − c) ≤ abc (d) [EC] 8abc ≥ (a + b)(b + c)(c + a) (e) [AP] 3(a + b)(b + c)(c + a) ≤ 8(a3 + b3 + c3 ) (f ) [MC] 2(a + b + c)(a2 + b2 + c2 ) ≥ 3(a3 + b3 + c3 + 3abc) (g) abc < a2 (s − a) + b2 (s − b) + c2 (s − c) ≤ 23 abc (h) bc(b + c) + ca(c + a) + ab(a + b) ≥ 48(s − a)(s − b)(s − c) 1 1 1 (i) s−a + s−b + s−c ≥ 9s a b c 3 + c+a + a+b <2 (j) [AMN], [MP] 2 ≤ b+c 15 s+a s+b s+c 9 (k) 4 ≤ b+c + c+a + a+b < 2 (l) [SR2] (a + b + c)3 ≤ 5[ab(a + b) + bc(b + c) + ca(c + a)] − 3abc Exercise 25. ([RS], R. Sondat) Let R, r, s be positive real numbers. Show that a necessary and sufficient condition for the existence of a triangle with circumradius R, inradius r, and semiperimeter s is s4 − 2(2R2 + 10Rr − r2 )s2 + r(4R + r)2 ≤ 0. √ Exercise 26. With the usual notation for a triangle, show that 4R + r ≥ 3s. 16 Exercise 27. ([WJB2],[RAS], W. J. Blundon) Let R and r denote the radii of the circumcircle and incircle of the triangle ABC. Let s be the semiperimeter of ABC. Show that √ s ≥ 2R + (3 3 − 4)r. Exercise 28. Let G and I be the centroid and incenter of the triangle ABC with inradius r, semiperimeter s, circumradius R. Show that ¢ 1¡ 2 GI 2 = s + 5r2 − 16Rr .17 9 Exercise 29. Show that, for any triangle with sides a, b, c, 2> 16 It’s 17 See a b c + + . b+c c+a a+b equivalent to the Hadwiger-Finsler inequality. the exercise 6. For a solution, see [KWL]. 25 Chapter 3 Homogenizations 3.1 Homogeneous Polynomial Inequalities Many inequality problems come with constraints such as ab = 1, xyz = 1, x + y + z = 1. A non-homogeneous symmetric inequality can be transformed into a homogeneous one. Then we apply two powerful theorems : Shur’s inequality and Muirhead’s theorem. We begin with a simple example. Problem 13. (Hungary 1996) Let a and b be positive real numbers with a + b = 1. Prove that a2 b2 1 + ≥ . a+1 b+1 3 Solution. Using the condition a + b = 1, we can reduce the given inequality to homogeneous one, i. e., 1 a2 b2 ≤ + or a2 b + ab2 ≤ a3 + b3 , 3 (a + b)(a + (a + b)) (a + b)(b + (a + b)) which follows from (a3 +b3 )−(a2 b+ab2 ) = (a−b)2 (a+b) ≥ 0. The equality holds if and only if a = b = 21 . The above inequality a2 b + ab2 ≤ a3 + b3 can be generalized as following : Theorem 9. Let a1 , a2 , b1 , b2 be positive real numbers such that a1 + a2 = b1 + b2 and max(a1 , a2 ) ≥ max(b1 , b2 ). Let x and y be nonnegative real numbers. Then, we have xa1 y a2 + xa2 y a1 ≥ xb1 y b2 + xb2 y b1 . Proof. Without loss of generality, we can assume that a1 ≥ a2 , b1 ≥ b2 , a1 ≥ b1 . If x or y is zero, then it clearly holds. So, we also assume that both x and y are nonzero. It’s easy to check ¡ ¢ xa1 y a2 + xa2 y a1 − xb1 y b2 − xb2 y b1 = xa2 y a2 xa1 −a2 + y a1 −a2 − xb1 −a2 y b2 −a2 − xb2 −a2 y b1 −a2 ¡ ¢¡ ¢ = xa2 y a2 xb1 −a2 − y b1 −a2 xb2 −a2 − y b2 −a2 ¢ ¡ b2 ¢ 1 ¡ b1 b1 b2 x − y x − y ≥ 0. = a a x 2y 2 Remark 1. When does the equality hold in the theorem 8? P P We now introduce two summation notations cyclic and sym . Let P (x, y, z) be a three variables function of x, y, z. Let us define : X P (x, y, z) = P (x, y, z) + P (y, z, x) + P (z, x, y), cyclic X P (x, y, z) = P (x, y, z) + P (x, z, y) + P (y, x, z) + P (y, z, x) + P (z, x, y) + P (z, y, x) sym 26 For example, we know that X X x3 y = x3 y + y 3 z + z 3 x, sym cyclic X x3 = 2(x3 + y 3 + z 3 ) X x2 y = x2 y + x2 z + y 2 z + y 2 x + z 2 x + z 2 y, sym xyz = 6xyz. sym Problem 14. (IMO 1984/1) Let x, y, z be nonnegative real numbers such that x + y + z = 1. Prove that 7 0 ≤ xy + yz + zx − 2xyz ≤ 27 . Solution. Using the condition x + y + z = 1, we reduce the given inequality to homogeneous one, i. e., 7 (x + y + z)3 . 27 P The left hand side inequality is trivial because it’s equivalent to 0 ≤ xyz + sym x2 y. The right hand side P P inequality simplifies to 7 cyclic x3 + 15xyz − 6 sym x2 y ≥ 0. In the view of 0 ≤ (xy + yz + zx)(x + y + z) − 2xyz ≤  7 X x3 + 15xyz − 6 it’s enough to show that 2 X cyclic x3 − X sym P cyclic x2 y =  X x2 y = 2 sym cyclic 2 X cyclic x3 ≥ X x3 − P sym X cyclic X x2 y  + 5 3xyz + cyclic (x2 y + xy 2 ) = x3 ≥ X cyclic x(x − y)(x − z) ≥ 0, cyclic which is a particular case of the theorem 10 in the next section. 27 X cyclic P cyclic The second inequality can be rewritten as X  sym x2 y and 3xyz + (x3 + y 3 ) −  P sym x3 − X x2 y  , sym x2 y. Note that (x3 + y 3 − x2 y − xy 2 ) ≥ 0. 3.2 Schur’s Theorem Theorem 10. (Schur) Let x, y, z be nonnegative real numbers. For any r > 0, we have X xr (x − y)(x − z) ≥ 0. cyclic Proof. Since the inequality is symmetric in the three variables, we may assume without loss of generality that x ≥ y ≥ z. Then the given inequality may be rewritten as (x − y)[xr (x − z) − y r (y − z)] + z r (x − z)(y − z) ≥ 0, and every term on the left-hand side is clearly nonnegative. Remark 2. When does the equality hold in Theorem 10? The following special case of Schur’s inequality is useful : X X X X X X x(x − y)(x − z) ≥ 0 ⇔ 3xyz + x3 ≥ x2 y ⇔ xyz + x3 ≥ 2 x2 y. cyclic sym cyclic sym sym sym Exercise 30. ([TZ], pp.142) Prove that for any acute triangle ABC, cot3 A + cot3 B + cot3 C + 6 cot A cot B cot C ≥ cot A + cot B + cot C. Exercise 31. (Korea 1998) Let I be the incenter of a triangle ABC. Prove that BC 2 + CA2 + AB 2 . 3 IA2 + IB 2 + IC 2 ≥ Exercise 32. ([IN], pp.103) Let a, b, c be the lengths of a triangle. Prove that a2 b + a2 c + b2 c + b2 a + c2 a + c2 b > a3 + b3 + c3 + 2abc. We present another solution of the problem 1 : (IMO 2000/2) Let a, b, c be positive numbers such that abc = 1. Prove that ¶µ ¶µ ¶ µ 1 1 1 b−1+ c−1+ ≤ 1. a−1+ b c a Second Solution. It is equivalent to the following homogeneous inequality1 : µ ¶µ ¶µ ¶ (abc)2/3 (abc)2/3 (abc)2/3 1/3 1/3 1/3 a − (abc) + b − (abc) + c − (abc) + ≤ abc. b c a After the substitution a = x3 , b = y 3 , c = z 3 with x, y, z > 0, it becomes µ ¶µ ¶µ ¶ (xyz)2 (xyz)2 (xyz)2 3 3 3 x − xyz + y − xyz + z − xyz + ≤ x3 y 3 z 3 , y3 z3 x3 which simplifies to ¡ or ¢¡ ¢¡ ¢ x2 y − y 2 z + z 2 x y 2 z − z 2 x + x2 y z 2 x − x2 y + y 2 z ≤ x3 y 3 z 3 3×3 y 3 z 3 + X cyclic or X x6 y 3 ≥ 3(x2 y)(y 2 z)(z 2 x) + cyclic X (x2 y)3 ≥ cyclic X sym an alternative homogenization, see the problem 1 in the chapter 2. 28 x5 y 2 z 2 cyclic which is a special case of Schur’s inequality. 1 For X x4 y 4 z + (x2 y)2 (y 2 z) Here is another inequality problem with the constraint abc = 1. Problem 15. (Tournament of Towns 1997) Let a, b, c be positive numbers such that abc = 1. Prove that 1 1 1 + + ≤ 1. a+b+1 b+c+1 c+a+1 Solution. We can rewrite the given inequality as following : 1 1 1 1 + + ≤ . a + b + (abc)1/3 b + c + (abc)1/3 c + a + (abc)1/3 (abc)1/3 We make the substitution a = x3 , b = y 3 , c = z 3 with x, y, z > 0. Then, it becomes 1 1 1 1 + 3 + 3 ≤ x3 + y 3 + xyz y + z 3 + xyz z + x3 + xyz xyz which is equivalent to X xyz (x3 + y 3 + xyz)(y 3 + z 3 + xyz) ≤ (x3 + y 3 + xyz)(y 3 + z 3 + xyz)(z 3 + x3 + xyz) cyclic and hence to P sym x6 y 3 ≥ P sym x5 y 2 z 2 , which is a special case of theorem 11 in the next section. 29 3.3 Muirhead’s Theorem Theorem 11. (Muirhead) Let a1 , a2 , a3 , b1 , b2 , b3 be real numbers such that a1 ≥ a2 ≥ a3 ≥ 0, b1 ≥ b2 ≥ b3 ≥ 0, a1 ≥ b1 , a1 + a2 ≥ b1 + b2 , a1 + a2 + a3 = b1 + b2 + b3 .2 P P Let x, y, z be positive real numbers. Then, we have sym xa1 y a2 z a3 ≥ sym xb1 y b2 z b3 . Proof. Case 1. b1 ≥ a2 : It follows from a1 ≥ a1 + a2 − b1 and from a1 ≥ b1 that a1 ≥ max(a1 + a2 − b1 , b1 ) so that max(a1 , a2 ) = a1 ≥ max(a1 +a2 −b1 , b1 ). From a1 +a2 −b1 ≥ b1 +a3 −b1 = a3 and a1 +a2 −b1 ≥ b2 ≥ b3 , we have max(a1 + a2 − b1 , a3 ) ≥ max(b2 , b3 ). Apply the theorem 8 twice to obtain X X xa1 y a2 z a3 = z a3 (xa1 y a2 + xa2 y a1 ) sym cyclic ≥ X z a3 (xa1 +a2 −b1 y b1 + xb1 y a1 +a2 −b1 ) cyclic = X xb1 (y a1 +a2 −b1 z a3 + y a3 z a1 +a2 −b1 ) cyclic ≥ X xb1 (y b2 z b3 + y b3 z b2 ) cyclic = X xb1 y b2 z b3 . sym Case 2. b1 ≤ a2 : It follows from 3b1 ≥ b1 + b2 + b3 = a1 + a2 + a3 ≥ b1 + a2 + a3 that b1 ≥ a2 + a3 − b1 and that a1 ≥ a2 ≥ b1 ≥ a2 + a3 − b1 . Therefore, we have max(a2 , a3 ) ≥ max(b1 , a2 + a3 − b1 ) and max(a1 , a2 + a3 − b1 ) ≥ max(b2 , b3 ). Apply the theorem 8 twice to obtain X X x a1 y a2 z a3 = xa1 (y a2 z a3 + y a3 z a2 ) sym cyclic ≥ X xa1 (y b1 z a2 +a3 −b1 + y a2 +a3 −b1 z b1 ) cyclic = X y b1 (xa1 z a2 +a3 −b1 + xa2 +a3 −b1 z a1 ) cyclic ≥ X y b1 (xb2 z b3 + xb3 z b2 ) cyclic = X xb1 y b2 z b3 . sym Remark 3. The equality holds if and only if x = y = z. However, if we allow x = 0 or y = 0 or z = 0, then one may easily check that the equality holds if and only if 3 x = y = z or x = y, z = 0 or y = z, x = 0 or z = x, y = 0. We can use Muirhead’s theorem to prove Nesbitt’s inequality. (Nesbitt) For all positive real numbers a, b, c, we have a b c 3 + + ≥ . b+c c+a a+b 2 2 Note the equality in the final equation. in this case, we assume that 00 = 1 in the sense that limx→0+ x0 = 1. In general, 00 is not defined. Note also that limx→0+ 0x = 0. 3 However, 30 Proof 6. Clearing the denominators of the inequality, it becomes X X X a(a + b)(a + c) ≥ 3(a + b)(b + c)(c + a) or a3 ≥ a2 b. 2 sym cyclic sym Problem 16. ((IMO 1995) Let a, b, c be positive numbers such that abc = 1. Prove that 1 1 1 3 + + ≥ . a3 (b + c) b3 (c + a) c3 (a + b) 2 Solution. It’s equivalent to 1 1 3 1 + + ≥ . a3 (b + c) b3 (c + a) c3 (a + b) 2(abc)4/3 P 1 Set a = x3 , b = y 3 , c = z 3 with x, y, z > 0. Then, it becomes cyclic x9 (y 3 +z 3 ) ≥ denominators, this becomes X X X X x12 y 9 z 3 + x9 y 9 z 6 ≥ 3 x11 y 8 z 5 + 6×8 y 8 z 8 x12 y 12 + 2 sym sym sym 3 2×4 y 4 z 4 . Clearing sym or à X X x12 y 12 − sym ! x11 y 8 z 5 à +2 sym X x12 y 9 z 3 − sym X ! x11 y 8 z 5 + à X sym x9 y 9 z 6 − sym X ! x8 y 8 z 8 ≥ 0, sym and every term on the left hand side is nonnegative by Muirhead’s theorem. We can also attack problem 10 and problem 11 with Schur’s inequality and Muirhead’s theorem. (Iran 1996) Let x, y, z be positive real numbers. Prove that µ ¶ 1 1 1 9 (xy + yz + zx) + + ≥ . (x + y)2 (y + z)2 (z + x)2 4 Second Solution. It’s equivalent to X X X X X 4 x5 y + 2 x4 yz + 6×2 y 2 z 2 − x4 y 2 − 6 x3 y 3 − 2 x3 y 2 z ≥ 0. sym sym cyclic sym cyclic We rewrite this as following à X x5 y − sym X ! x4 y 2 sym à +3 X x5 y − sym X ! x3 y 3  + 2xyz  sym  X x(x − y)(x − z) ≥ 0. cyclic By Muirhead’s theorem and Schur’s inequality, it’s a sum of three terms which are nonnegative. Let x, y, z be nonnegative real numbers with xy + yz + zx = 1. Prove that 1 1 5 1 + + ≥ . x+y y+z z+x 2 Second Solution. Using xy + yz + zx = 1, we homogenize the given inequality as following : µ ¶2 µ ¶2 1 1 1 5 (xy + yz + zx) + + ≥ x+y y+z z+x 2 or 4 X sym x5 y + X sym x4 yz + 14 X x3 y 2 z + 38×2 y 2 z 2 ≥ X sym sym 31 x4 y 2 + 3 X sym x3 y 3 or à X x5 y − sym X ! x4 y 2 sym à +3 X x5 y − sym X ! x3 y 3 + xyz sym à X x3 + 14 sym X ! x2 y + 38xyz ≥ 0. sym By Muirhead’s theorem, we get the result. In the above inequality, without the condition xy + yz + zx = 1, the equality holds if and only if x = y, z = 0 or y = z, x = 0 or z = x, y = 0. Since xy + yz + zx = 1, the equality occurs when (x, y, z) = (1, 1, 0), (1, 0, 1), (0, 1, 1). Now, we apply Muirhead’s theorem to obtain a geometric inequality [ZsJc] : Problem 17. If ma ,mb ,mc are medians and ra ,rb ,rc the exradii of a triangle, prove that rb rc rc ra ra rb + + ≥ 3. ma mb mb mc mc ma An Impossible Verification. Let 2s = a + b + c. Using the well-known identities r s(s − b)(s − c) 1p 2 ra = , ma = 2b + 2c2 − a2 , etc. s−a 2 we have X rb rc X 4s(s − a) p = . 2 2 mb mc (2c + 2a − b2 )(2a2 + 2b2 − c2 ) cyclic cyclic Applying the AM-GM inequality, we obtain X rb rc X X 2(a + b + c)(b + c − a) 8s(s − a) ≥ = . 2 2 2 2 2 2 mb mc (2c + 2a − b ) + (2a + 2b − c ) 4a2 + b2 + c2 cyclic cyclic cyclic We now give a moonshine proof of the inequality X 2(a + b + c)(b + c − a) ≥ 3. 4a2 + b2 + c2 cyclic After expanding the above inequality, it becomes X X X X X X 2 a6 + 4 a4 bc + 20 a3 b2 c + 68 a3 b3 + 16 a5 b ≥ 276a2 b2 c2 + 27 a4 b2 . cyclic cyclic sym cyclic cyclic cyclic We see that this cannot be directly proven by applying Muirhead’s theorem. Since a, b, c are the sides of a triangle, we can make the Ravi Substitution a = y + z, b = z + x, c = x + y, where x, y, z > 0. After some brute-force algebra, we can rewrite the above inequality as X X X X X 25 x6 + 230 x5 y + 115 x4 y 2 + 10 x3 y 3 + 80 x4 yz sym sym ≥ 336 sym X x3 y 2 z + 124 sym X sym sym Now, by Muirhead’s theorem, we get the result ! 32 x2 y 2 z 2 . sym 3.4 Polynomial Inequalities with Degree 3 The solution of problem 13 shows us difficulties in applying Muirhead’s theorem. Furthermore, there exist homogeneous symmetric polynomial inequalities which cannot be verified by just applying Muirhead’s theorem. See the following inequality : X X X X X X 5 x6 + 15 x4 y 2 + 2 x3 y 2 z + 3×2 y 2 z 2 ≥ 8 x5 y + 8 x4 yz + 16 x3 y 3 sym cyclic sym sym cyclic cyclic This holds for all positive real numbers z. However, it is not a direct consequence of Muirhead’s P x, y, and P theorem because the coefficients of sym x5 y and cyclic x3 y 3 are too big. In fact, it is equivalent to 1 X (y − z)4 (x2 + 15y 2 + 15z 2 + 8xy + 4yz + 8zx) ≥ 0.4 6 cyclic Another example is 1 X 4 3 X 2 2 X 3 x + x y ≥ x y. 2 2 sym cyclic cyclic We realized that the above inequality is stronger than X X X X x2 (x − y)(x − z) ≥ 0 or x4 + x2 y 2 ≥ x3 y. cyclic cyclic cyclic sym It can be proved by the identities   X X X 3 1 x4 + x2 y 2 − x3 y  = (x − y)4 + (y − z)4 + (z − x)4 4 2 2 sym cyclic or cyclic   X X X 3 1 x4 + x2 y 2 − 2 x3 y  = (x2 + y 2 + z 2 − xy − yz − zx)2 . 2 2 sym cyclic cyclic As I know, there is no general criterion to attack the symmetric polynomial inequalities. However, there is a result for the homogeneous symmetric polynomial inequalities with degree 3. It’s a direct consequence of Muirhead’s theorem and Schur’s inequality. Theorem 12. Let P (u, v, w) ∈ R[u, v, w] be a homogeneous symmetric polynomial with degree 3. Then the following two statements are equivalent. (a) P (1, 1, 1), P (1, 1, 0), P (1, 0, 0) ≥ 0. (b) P (x, y, z) ≥ 0 for all x, y, z ≥ 0. Proof. (See [SR].) We only prove that (a) implies (b). Let X X P (u, v, w) = A u3 + B u2 v + Cuvw. sym cyclic Let p = P (1, 1, 1) = 3A + 6B + C, q = P (1, 1, 0) = A + B, and r = P (1, 0, 0) = A. We have A = r, B = q − r, C = p − 6q + 3r, and p, q, r ≥ 0. Let x, y, z ≥ 0. It follows that X X x2 y + (p − 6q + 3r)xyz. x3 + (q − r) P (x, y, z) = r sym cyclic However, we see that   P (x, y, z) = r  X cyclic 4 See x3 + 3xyz − X x2 y  + q sym à X sym [JC]. 33 ! x2 y − 6xyz + pxyz ≥ 0. Here is an alternative way to prove of the fact that P (x, y, z) ≥ 0. Case 1. q ≥ r : We find that r P (x, y, z) = 2 à X 3 x − sym X ! xyz + (q − r) à X sym 2 x y− sym X ! xyz + pxyz. sym and that the every term on the right hand side is nonnegative. Case 2. q ≤ r : We find that q P (x, y, z) = 2 à X sym x3 − X  ! xyz + (r − q)  sym  X x3 + 3xyz − X x2 y  + pxyz. sym cyclic and that the every term on the right hand side is nonnegative. For example, we can apply the theorem 11 to check the inequality in the problem 14. (IMO 1984/1) Let x, y, z be nonnegative real numbers such that x + y + z = 1. Prove that 7 0 ≤ xy + yz + zx − 2xyz ≤ 27 . Solution. Using x + y + z = 1, we homogenize the given inequality as following : 0 ≤ (xy + yz + zx)(x + y + z) − 2xyz ≤ 7 (x + y + z)3 27 Let us define L(u, v, w), R(u, v, w) ∈ R[u, v, w] by L(u, v, w) = (uv + vw + wu)(u + v + w) − 2uvw, R(u, v, w) = 7 (u + v + w)3 − (uv + vw + wu)(u + v + w) + 2uvw. 27 However, one may easily check that L(1, 1, 1) = 7, L(1, 1, 0) = 2, L(1, 0, 0) = 0, R(1, 1, 1) = 0, R(1, 1, 0) = 7 and R(1, 0, 0) = 27 . 2 27 , Exercise 33. (M. S. Klamkin [MEK2]) Determine the maximum and minimum values of x2 + y 2 + z 2 + λxyz where x + y + z = 1, x, y, z ≥ 0, and λ is a given constant. Exercise 34. (Walter Janous [MC]) let x, y, z ≥ 0 with x + y + z = 1. For fixed real numbers a ≥ 0 and b, determine the maximum c = c(a, b) such that a + bxyz ≥ c(xy + yz + zx). Here is the criterion for homogeneous symmetric polynomial inequalities for the triangles : Theorem 13. (K. B. Stolarsky) Let P (u, v, w) be a real symmetric form of degree 3.5 If P (1, 1, 1), P (1, 1, 0), P (2, 1, 1) ≥ 0, then we have P (a, b, c) ≥ 0, where a, b, c are the lengths of the sides of a triangle. Proof. Make the Ravi substitution a = y + z, b = z + x, c = x + y and apply the above theorem. We leave the details for the readers. For an alternative proof, see [KBS]. As noted in [KBS], we can apply Stolarsky’s theorem to prove cubic inequalities in triangle geometry. We recall the exercise 11. 5 P (x, y, z) = P sym ¡ 3 ¢ px + qx2 y + rxyz (p, q, r ∈ R.) 34 Let a, b, c be the lengths of the sides of a triangle. Let s be the semi-perimeter of the triangle. Then, the following inequalities holds. 2 (a) 3(ab + bc + ca) ≤ (a + b + c) ¡ 2< 4(ab ¢+ bc + ca) 36 2 2 2 (b) [JfdWm] a + b + c ≥ 35 s + abc s (c) [AP] 8(s − a)(s − b)(s − c) ≤ abc (d) [EC] 8abc ≥ (a + b)(b + c)(c + a) (e) [AP] 3(a + b)(b + c)(c + a) ≤ 8(a3 + b3 + c3 ) (f) [MC] 2(a + b + c)(a2 + b2 + c2 ) ≥ 3(a3 + b3 + c3 + 3abc) (g) abc < a2 (s − a) + b2 (s − b) + c2 (s − c) ≤ 23 abc (h) bc(b + c) + ca(c + a) + ab(a + b) ≥ 48(s − a)(s − b)(s − c) 1 1 1 (i) s−a + s−b + s−c ≥ 9s a b c 3 + c+a + a+b <2 (j) [AMN], [MP] 2 ≤ b+c 15 s+a s+b s+c 9 (k) 4 ≤ b+c + c+a + a+b < 2 (l) [SR] (a + b + c)3 ≤ 5[ab(a + b) + bc(b + c) + ca(c + a)] − 3abc Proof. For example, we show the right hand side inequality in (j). It’s equivalent to the cubic inequality T (a, b, c) ≥ 0, where µ ¶ b c a + + . T (a, b, c) = 2(a + b)(b + c)(c + a) − (a + b)(b + c)(c + a) b+c c+a a+b Since T (1, 1, 1) = 4, T (1, 1, 0) = 0, and T (2, 1, 1) = 6, the result follows from Stolarsky’s theorem. For alternative proofs of the above 12 inequalities, see [GI]. 35 3.5 Supplementary Problems for Chapter 3 Exercise 35. Let x, y, z be positive real numbers. Prove that (x + y − z)(x − y)2 + (y + z − x)(y − z)2 + (z + x − y)(z − x)2 ≥ 0. Exercise 36. Let x, y, z be positive real numbers. Prove that (x2 + y 2 − z 2 )(x − y)2 + (y 2 + z 2 − x2 )(y − z)2 + (z 2 + x2 − y 2 )(z − x)2 ≥ 0. Exercise 37. (APMO 1998) Let a, b, c be positive real numbers. Prove that ¶ µ µ ¶ ³ c´ a´ b ³ a+b+c 1+ 1+ 1+ ≥2 1+ √ . 3 b c a abc Exercise 38. (Ireland 2000) Let x, y ≥ 0 with x + y = 2. Prove that x2 y 2 (x2 + y 2 ) ≤ 2. Exercise 39. (IMO Short-listed 1998) Let x, y, z be positive real numbers such that xyz = 1. Prove that y3 z3 3 x3 + + ≥ . (1 + y)(1 + z) (1 + z)(1 + x) (1 + x)(1 + y) 4 Exercise 40. (United Kingdom 1999) Some three nonnegative real numbers p, q, r satisfy p + q + r = 1. Prove that 7(pq + qr + rp) ≤ 2 + 9pqr. 36 Chapter 4 Normalizations 4.1 Normalizations In the previous chapter, we transformed non-homogeneous inequalities into homogeneous ones. On the other hand, homogeneous inequalities also can be normalized in various ways. We offer two alternative solutions of the problem 8 by normalizations : (IMO 2001/2) Let a, b, c be positive real numbers. Prove that √ a b c +√ +√ ≥ 1. a2 + 8bc b2 + 8ca c2 + 8ab Second Solution. We make the substitution x = a a+b+c , y= b a+b+c , z= c 1 a+b+c . The problem is xf (x2 + 8yz) + yf (y 2 + 8zx) + zf (z 2 + 8xy) ≥ 1, where f (t) = √1t . Since the function f is convex down on R+ and x + y + z = 1, we apply (the weighted) Jensen’s inequality to have xf (x2 + 8yz) + yf (y 2 + 8zx) + zf (z 2 + 8xy) ≥ f (x(x2 + 8yz) + y(y 2 + 8zx) + z(z 2 + 8xy)). Note that f (1) = 1. Since the function f is strictly decreasing, it suffices to show that 1 ≥ x(x2 + 8yz) + y(y 2 + 8zx) + z(z 2 + 8xy). Using x + y + z = 1, we homogenize it as (x + y + z)3 ≥ x(x2 + 8yz) + y(y 2 + 8zx) + z(z 2 + 8xy). However, this is easily seen from (x + y + z)3 − x(x2 + 8yz) − y(y 2 + 8zx) − z(z 2 + 8xy) = 3[x(y − z)2 + y(z − x)2 + z(x − y)2 ] ≥ 0. In the above solution, we normalized to x + y + z = 1. We now prove it by normalizing to xyz = 1. ab Third Solution. We make the substitution x = abc2 , y = ca b2 , z = c2 . Then, we get xyz = 1 and the inequality becomes 1 1 1 √ +√ +√ ≥1 1 + 8y 1 + 8x 1 + 8z which is equivalent to X p (1 + 8x)(1 + 8y) ≥ p (1 + 8x)(1 + 8y)(1 + 8z) cyclic 1 Dividing by a + b + c gives the equivalent inequality P cyclic 37 r a a+b+c 8bc a2 + (a+b+c)2 (a+b+c)2 ≥ 1. hence, after squaring both sides, equivalent to X √ p 8(x + y + z) + 2 (1 + 8x)(1 + 8y)(1 + 8z) 1 + 8x ≥ 510. cyclic Recall that xyz = 1. The AM-GM inequality gives us x + y + z ≥ 3, 8 8 8 (1 + 8x)(1 + 8y)(1 + 8z) ≥ 9x 9 · 9y 9 · 9z 9 = 729 and X √ 1 + 8x ≥ cyclic X p 8 4 9x 9 ≥ 9(xyz) 27 = 9. cyclic Using these three inequalities, we get the result. We now present another proofs of Nesbitt’s inequality. (Nesbitt) For all positive real numbers a, b, c, we have a b c 3 + + ≥ . b+c c+a a+b 2 Proof 7. We may normalize to a + b + c = 1. Note that 0 < a, b, c < 1. The problem is now to prove X cyclic X 3 x a = f (a) ≥ , where f (x) = . b+c 2 1−x cyclic Since f is concave down on (0, 1), Jensen’s inequality shows that µ ¶ µ ¶ X a+b+c 1 X 1 1 3 f (a) ≥ f =f = or f (a) ≥ . 3 3 3 2 2 cyclic cyclic Proof 8. As in the previous proof, we need to prove X cyclic a 3 ≥ , where a + b + c = 1. 1−a 2 It follows from 4x − (1 − x)(9x − 1) = (3x − 1)2 or 4x ≥ (1 − x)(9x − 1) that X cyclic X 9a − 1 3 a 9 X 3 a− = . ≥ = 1−a 4 4 4 2 cyclic cyclic 38 4.2 Classical Theorems : Cauchy-Schwartz, (Weighted) AM-GM, and Hölder We now illustrate the normalization technique to establish classical theorems. Theorem 14. (The Cauchy-Schwartz inequality) Let a1 , · · · , an , b1 , · · · , bn be real numbers. Then, we have (a1 2 + · · · + an 2 )(b1 2 + · · · + bn 2 ) ≥ (a1 b1 + · · · + an bn )2 . p √ Proof. Let A = a1 2 + · · · + an 2 and B = b1 2 + · · · + bn 2 . In the case when A = 0, we get a1 = · · · = an = 0. Thus, the given inequality clearly holds. So, we now may assume that A, B > 0. Now, we make the substitution xi = aAi (i = 1, · · · , n). Then, it’s equivalent to (x1 2 + · · · + xn 2 )(b1 2 + · · · + bn 2 ) ≥ (x1 b1 + · · · + xn bn )2 . However, we have x1 2 + · · · + xn 2 = 1. (Why?). Hence, it’s equivalent to b1 2 + · · · + bn 2 ≥ (x1 b1 + · · · + xn bn )2 . Next, we make the substitution yi = bi B (i = 1, · · · , n). Then, it’s equivalent to 1 = y1 2 + · · · + yn 2 ≥ (x1 y1 + · · · + xn yn )2 or 1 ≥ |x1 y1 + · · · + xn yn |. Hence, we need to to show that |x1 y1 + · · · + xn yn | ≤ 1, where x1 2 + · · · + xn 2 = y1 2 + · · · + yn 2 = 1. However, it’s very easy. We apply the AM-GM inequality to deduce |x1 y1 + · · · + xn yn | ≤ |x1 y1 | + · · · + |xn yn | ≤ xn 2 + yn 2 A+B x1 2 + y1 2 + ··· + = = 1. 2 2 2 Exercise 41. Prove the Lagrange’s identity : n X i=1 ai 2 n X i=1 à 2 bi − n X !2 ai bi X = i=1 2 (ai bj − aj bi ) . 1≤i 1 such that x1 + y1 + z1 = 2, p √ √ √ x + y + z ≥ x − 1 + y − 1 + z − 1. Second Solution. We notice that 1 1 1 x−1 y−1 z−1 + + =2 ⇔ + + = 1. x y z x y z We now apply the Cauchy-Schwartz inequality to deduce s µ ¶ p √ √ √ x−1 y−1 z−1 x + y + z = (x + y + z) + + ≥ x − 1 + y − 1 + z − 1. x y z Problem 18. (Gazeta Matematicã, Hojoo Lee) Prove that, for all a, b, c > 0, p p p p p p a4 + a2 b2 + b4 + b4 + b2 c2 + c4 + c4 + c2 a2 + a4 ≥ a 2a2 + bc + b 2b2 + ca + c 2c2 + ab. Solution. We obtain the chain of equalities and inequalities s ¶ µ ¶ X p X µ a2 b2 a2 b2 4 2 2 4 4 4 a +a b +b = a + + b + 2 2 cyclic cyclic Ãr ! r 2 b2 2 b2 a a 1 X a4 + ≥ √ + b4 + (Cauchy − Schwartz) 2 2 2 cyclic Ãr ! r 2 b2 2 c2 1 X a a = √ a4 + + a4 + 2 2 2 cyclic sµ ¶µ ¶ √ X 4 a2 c2 a2 b2 4 4 2 a + (AM − GM) ≥ a + 2 2 cyclic r √ X a2 bc ≥ 2 a4 + (Cauchy − Schwartz) 2 cyclic X p = 2a4 + a2 bc . cyclic 40 Using the same idea in the proof of the Cauchy-Schwartz inequality, we find a natural generalization : Theorem 15. Let aij (i, j = 1, · · · , n) be positive real numbers. Then, we have (a11 n + · · · + a1n n ) · · · (an1 n + · · · + ann n ) ≥ (a11 a21 · · · an1 + · · · + a1n a2n · · · ann )n . Proof. Since the inequality is homogeneous, as in the proof of the theorem 11, we can normalize to 1 (ai1 n + · · · + ain n ) n = 1 or ai1 n + · · · + ain n = 1 (i = 1, · · · , n). Pn Then, the inequality takes the form a11 a21 · · · an1 + · · · + a1n a2n · · · ann ≤ 1 or i=1 ai1 · · · ain ≤ 1. Hence, it suffices to show that, for all i = 1, · · · , n, ai1 · · · ain ≤ 1 , where ai1 + · · · + ain = 1. n To finish the proof, it remains to show the following homogeneous inequality : Theorem 16. (AM-GM inequality) Let a1 , · · · , an be positive real numbers. Then, we have 1 a1 + · · · + an ≥ (a1 · · · an ) n . n Proof. Since it’s homogeneous, we may rescale a1 , · · · , an so that a1 · · · an = 1. that a1 · · · an = 1 =⇒ a1 + · · · + an ≥ n. 2 Hence, we want to show √ The proof is by induction on n. If n = 1, it’s trivial. If n = 2, then we get a1 + a2 − 2 = a1 + a2 − 2 a1 a2 = √ √ 2 ( a1 − a2 ) ≥ 0. Now, we assume that it holds for some positive integer n ≥ 2. And let a1 , · · · , an+1 be positive numbers such that a1 · · · an an+1 =1. We may assume that a1 ≥ 1 ≥ a2 . (Why?) Since (a1 a2 )a3 · · · an = 1, by the induction hypothesis, we have a1 a2 + a3 + · · · + an+1 ≥ n. Thus, it suffices to show that a1 a2 + 1 ≤ a1 + a2 . However, we have a1 a2 + 1 − a1 − a2 = (a1 − 1)(a2 − 1) ≤ 0. The following simple observation is not tricky : Let a, b > 0 and m, n ∈ N. Take x1 = · · · = xm = a and xm+1 = · · · = xxm+n = b. Applying the AM-GM inequality to x1 , · · · , xm+n > 0, we obtain 1 m n ma + nb m n ≥ (am bn ) m+n or a+ b ≥ a m+n b m+n . m+n m+n m+n Hence, for all positive rationals ω1 and ω2 with ω1 + ω2 = 1, we get ω1 a + ω2 b ≥ a ω1 b ω2 . We immediately have Theorem 17. Let ω1 , ω2 > 0 with ω1 + ω2 = 1. Then, for all x, y > 0, we have ω1 x + ω2 y ≥ x ω1 y ω2 . Proof. We can choose a positive rational sequence a1 , a2 , a3 , · · · such that lim an = ω1 . n→∞ And letting bi = 1 − ai , we get lim bn = ω2 . n→∞ From the previous observation, we have an x + bn y ≥ xan y bn Now, taking the limits to both sides, we get the result. 2 Set xi = ai 1 (a1 ···an ) n (i = 1, · · · , n). Then, we get x1 · · · xn = 1 and it becomes x1 + · · · + xn ≥ n. 41 Modifying slightly the above arguments, we obtain Theorem 18. (Weighted AM-GM inequality) Let ω1 , · · · , ωn be positive real numbers satisfying ω1 + · · · + ωn = 1. Then, for all x1 , · · · , xn > 0, we have ω1 x1 + · · · + ωn xn ≥ x1 ω1 · · · xn ωn . Recall that the AM-GM inequality is used to deduce the theorem 12, which is a generalization of the Cauchy-Schwartz inequality. Since we now get the weighted version of the AM-GM inequality, we establish weighted version of the Cauchy-Schwartz inequality. It’s called Hölder’s Inequality : Theorem 19. (Hölder) Let xij (i = 1, · · · , m, j = 1, · · · n) be positive real numbers. Suppose that ω1 , · · · , ωn are positive real numbers satisfying ω1 + · · · + ωn = 1. Then, we have Ãm !ωj n m Y n Y X X xij ≥ xij ωj . j=1 i=1 i=1 j=1 Proof. Since the inequality is homogeneous, as in the proof of the theorem 12, we may rescale x1j , · · · , xmj so that x1j + · · · + xmj = 1 for each j ∈ {1, · · · , n}. Then, we need to show that n Y 1ωj ≥ j=1 m Y n X xij ωj or 1 ≥ i=1 j=1 m Y n X xij ωj . i=1 j=1 The weighted AM-GM inequality provides that n X j=1 ωj xij ≥ n Y xij ωj (i ∈ {1, · · · , m}) =⇒ j=1 m X n X ωj xij ≥ i=1 j=1 m Y n X i=1 j=1 However, we immediately have m X n X i=1 j=1 ωj xij = n X m X j=1 i=1 ωj xij = n X j=1 42 ωj Ãm X i=1 ! xij = n X j=1 ωj = 1. xij ωj . 4.3 Homogenizations and Normalizations Here, we present an inequality problem which is solved by the techniques we studied : normalization and homogenization. Problem 19. (IMO 1999/2) Let n be an integer with n ≥ 2. (a) Determine the least constant C such that the inequality 4  X X xi xj (x2i + x2j ) ≤ C  xi  1≤i 0. Since the inequality is homogeneous, we may normalize to x1 + · · · + xn = 1. We denote X F (x1 , · · · , xn ) = xi xj (x2i + x2j ). 1≤i 1 be a positive integer and a1 , · · · , an , b1 , · · · , bn be positive real numbers. Prove the following inequality.  2 X X X  ai bj  ≥ ai aj bi bj . i6=j i6=j i6=j M 5. (C2176, Sefket Arslanagic) Prove that 1 1 1 ((a1 + b1 ) · · · (an + bn )) n ≥ (a1 · · · an ) n + (b1 · · · bn ) n where a1 , · · · , an , b1 , · · · , bn > 0 M 6. (Korea 2001) Let x1 , · · · , xn and y1 , · · · , yn be real numbers satisfying x1 2 + · · · + xn 2 = y1 2 + · · · + yn 2 = 1 Show that ¯ ¯ n ¯ ¯ X ¯ ¯ 2 ¯1 − xi yi ¯ ≥ (x1 y2 − x2 y1 )2 ¯ ¯ i=1 and determine when equality holds. 45 M 7. (Singapore 2001) Let a1 , · · · , an , b1 , · · · , bn be real numbers between 1001 and 2002 inclusive. Suppose that n n X X ai 2 = bi 2 . i=1 Prove that n X ai 3 i=1 bi i=1 n ≤ 17 X 2 ai . 10 i=1 Determine when equality holds. M 8. ([EWW-AI], Abel’s inequality) Let a1 , · · · , aN , x1 , · · · , xN be real numbers with xn ≥ xn+1 > 0 for all n. Show that |a1 x1 + · · · + aN xN | ≤ Ax1 where A = max{|a1 |, |a1 + a2 |, · · · , |a1 + · · · + aN |}. M 9. (China 1992) For every integer n ≥ 2 find the smallest positive number λ = λ(n) such that if 1 , b1 , · · · , bn > 0, a1 + · · · + an = b1 + · · · + bn = 1 2 0 ≤ a1 , · · · , an ≤ then b1 · · · bn ≤ λ(a1 b1 + · · · + an bn ). M 10. (C2551, Panos E. Tsaoussoglou) Suppose that a1 , · · · , an are positive real numbers. Let ej,k = n − 1 if j = k and ej,k = n − 2 otherwise. Let dj,k = 0 if j = k and dj,k = 1 otherwise. Prove that n Y n X ej,k ak 2 ≥ j=1 k=1 n Y j=1 à n X !2 dj,k ak k=1 M 11. (C2627, Walther Janous) Let x1 , · · · , xn (n ≥ 2) be positive real numbers and let x1 + · · · + xn . Let a1 , · · · , an be non-negative real numbers. Determine the optimum constant C(n) such that n X aj (sn − xj ) j=1 xj  ≥ C(n)  n Y  n1 aj  . j=1 M 12. (Hungary-Israel Binational Mathematical Competition 2000) Suppose that k and l are two given positive integers and aij (1 ≤ i ≤ k, 1 ≤ j ≤ l) are given positive numbers. Prove that if q ≥ p > 0, then    pq  p1  à k ! pq  q1 l k l X X X X    ≤  aij p aij q   .  j=1 i=1 i=1 j=1 M 13. ([EWW-KI] Kantorovich inequality) Suppose x1 < · · · < xn are given positive numbers. Let λ1 , · · · , λn ≥ 0 and λ1 + · · · + λn = 1. Prove that ! à n !à n X X λi A2 ≤ 2, λi x i x G i=1 i=1 i where A = x1 +xn 2 and G = √ x1 xn . M 14. (Czech-Slovak-Polish Match 2001) Let n ≥ 2 be an integer. Show that (a1 3 + 1)(a2 3 + 1) · · · (an 3 + 1) ≥ (a1 2 a2 + 1)(a2 2 a3 + 1) · · · (an 2 a1 + 1) for all nonnegative reals a1 , · · · , an . 46 M 15. (C1868, De-jun Zhao) Let n ≥ 3, a1 > a2 > · · · > an > 0, and p > q > 0. Show that a1 p a2 q + a2 p a3 q + · · · + an−1 p an q + an p a1 q ≥ a1 q a2 p + a2 q a3 p + · · · + an−1 q an p + an q a1 p M 16. (Baltic Way 1996) For which positive real numbers a, b does the inequality x1 x2 + x2 x3 + · · · + xn−1 xn + xn x1 ≥ x1 a x2 b x3 a + x2 a x3 b x4 a + · · · + xn a x1 b x2 a holds for all integers n > 2 and positive real numbers x1 , · · · , xn . M 17. (IMO short List 2000) Let x1 , x2 , · · · , xn be arbitrary real numbers. Prove the inequality √ x1 x2 xn + + ··· + < n. 1 + x1 2 1 + x1 2 + x2 2 1 + x1 2 + · · · + xn 2 M 18. (MM1479, Donald E. Knuth) Let Mn be the maximum value of the quantity xn x2 x1 + + ··· + (1 + x1 + · · · + xn )2 (1 + x2 + · · · + xn )2 (1 + xn )2 over all nonnegative real numbers (x1 , · · · , xn ). At what point(s) does the maximum occur ? Express Mn in terms of Mn−1 , and find limn→∞ Mn . M 19. (IMO 1971) Prove the following assertion is true for n = 3 and n = 5 and false for every other natural number n > 2 : if a1 , · · · , an are arbitrary real numbers, then n Y X (ai − aj ) ≥ 0. i=1 i6=j M 20. (IMO 2003) Let x1 ≤ x2 ≤ · · · ≤ xn be real numbers. (a) Prove that  2 X 2(n2 − 1) X  |xi − xj | ≤ (xi − xj )2 . 3 1≤i,j≤n 1≤i,j≤n (b) Show that the equality holds if and only if x1 , x2 , · · · , xn is an arithmetic sequence. M 21. (Bulgaria 1995) Let n ≥ 2 and 0 ≤ x1 , · · · , xn ≤ 1. Show that (x1 + x2 + · · · + xn ) − (x1 x2 + x2 x3 + · · · + xn x1 ) ≤ hni 2 , and determine when there is equality. M 22. (MM1407, Murry S. Klamkin) Determine the maximum value of the sum x1 p + x2 p + · · · + xn p − x1 q x2 r − x2 q x3 r − · · · xn q x1 r , where p, q, r are given numbers with p ≥ q ≥ r ≥ 0 and 0 ≤ xi ≤ 1 for all i. M 23. (IMO Short List 1998) Let a1 , a2 , · · · , an be positive real numbers such that a1 + a2 + · · · + an < 1. Prove that 1 a1 a2 · · · an (1 − (a1 + a2 + · · · + an )) ≤ n+1 . (a1 + a2 + · · · + an )(1 − a1 )(1 − a2 ) · · · (1 − an ) n M 24. (IMO Short List 1998) Let r1 , r2 , · · · , rn be real numbers greater than or equal to 1. Prove that 1 1 n + ··· + ≥ . 1 r1 + 1 rn + 1 (r1 · · · rn ) n + 1 47 M 25. (Baltic Way 1991) Prove that, for any real numbers a1 , · · · , an , X 1≤i,j≤n ai aj ≥ 0. i+j−1 M 26. (India 1995) Let x1 , x2 , · · · , xn be positive real numbers whose sum is 1. Prove that r x1 xn n + ··· + ≥ . 1 − x1 1 − xn n−1 M 27. (Turkey 1997) Given an integer n ≥ 2, Find the minimal value of x2 5 xn 5 x1 5 + + ··· x2 + x3 + · · · + xn x3 + · · · + xn + x1 x1 + x3 + · · · + xn−1 for positive real numbers x1 , · · · , xn subject to the condition x1 2 + · · · + xn 2 = 1. M 28. (China 1996) Suppose n ∈ N, x0 = 0, x1 , · · · , xn > 0, and x1 + · · · + xn = 1. Prove that 1≤ n X i=1 √ π xi √ < 2 1 + x0 + · · · + xi−1 xi + · · · + xn M 29. (Vietnam 1998) Let x1 , · · · , xn be positive real numbers satisfying 1 1 1 + ··· + = . x1 + 1998 xn + 1998 1998 Prove that 1 (x1 · · · xn ) n ≥ 1998 n−1 M 30. (C2768 Mohammed Aassila) Let x1 , · · · , xn be n positive real numbers. Prove that √ x2 xn n x1 +√ + ··· + √ ≥√ 2 2 2 x1 x2 + x2 x2 x3 + x3 xn x1 + x1 2 M 31. (C2842, George Tsintsifas) Let x1 , · · · , xn be positive real numbers. Prove that 1 (a) x1 n + · · · + xn n n(x1 · · · xn ) n + ≥ 2, nx1 · · · xn x1 + · · · + xn (b) x1 n + · · · + xn n (x1 · · · xn ) n + ≥ 1. x1 · · · xn x1 + · · · + xn 1 M 32. (C2423, Walther Janous) Let x1 , · · · , xn (n ≥ 2) be positive real numbers such that x1 +· · ·+xn = 1. Prove that µ ¶ µ ¶ µ ¶ µ ¶ 1 1 n − x1 n − xn 1+ ··· 1 + ≥ ··· x1 xn 1 − x1 1 − xn Determine the cases of equality. M 33. (C1851, Walther Janous) Let x1 , · · · , xn (n ≥ 2) be positive real numbers such that x1 2 + · · · + xn 2 = 1. Prove that √ √ n X 2 n−1 2 n+1 2 + xi √ ≤ √ ≤ . 5 + xi 5 n−1 5 n+1 i=1 48 M 34. (C1429, D. S. Mitirinovic, J. E. Pecaric) Show that n X i=1 xi ≤n−1 xi 2 + xi+1 xi+2 where x1 , · · · , xn are n ≥ 3 positive real numbers. Of course, xn+1 = x1 , xn+2 = x2 . 1 M 35. (Belarus 1998 S. Sobolevski) Let a1 ≤ a2 ≤ · · · ≤ an be positive real numbers. Prove the inequalities a1 a1 + · · · + an n , (a) 1 1 ≥ a · n n a1 + · · · + an (b) where k = 1 a1 n + ··· + 1 an ≥ 2k a1 + · · · + an , · 1 + k2 n an a1 . M 36. (Hong Kong 2000) Let a1 ≤ a2 ≤ · · · ≤ an be n real numbers such that a1 + a2 + · · · + an = 0. Show that a1 2 + a2 2 + · · · + an 2 + na1 an ≤ 0. M 37. (Poland 2001) Let n ≥ 2 be an integer. Show that µ ¶ X n n X n i xi + ≥ ixi 2 i=1 i=1 for all nonnegative reals x1 , · · · , xn . M 38. (Korea 1997) Let a1 , · · · , an be positive numbers, and define A= 1 a1 + · · · + an , G = (a1 · · ·n ) n , H = n (a) If n is even, show that A ≤ −1 + 2 H µ A G 1 a1 n + ··· + 1 an ¶n (b) If n is odd, show that n − 2 2(n − 1) A ≤− + H n n . µ A G ¶n . M 39. (Romania 1996) Let x1 , · · · , xn , xn+1 be positive reals such that xn+1 = x1 + · · · + xn . Prove that n p X xi (xn+1 − xi ) ≤ p xn+1 (xn+1 − xi ) i=1 M 40. (C2730, Peter Y. Woo) Let AM (x1 , · · · , xn ) and GM (x1 , · · · , xn ) denote the arithmetic mean and the geometric mean of the positive real numbers x1 , · · · , xn respectively. Given positive real numbers a1 , · · · , an , b1 , · · · , bn , (a) prove that GM (a1 + b1 , · · · , an + bn ) ≥ GM (a1 , · · · , an ) + GM (b1 , · · · , bn ). For each real number t ≥ 0, define f (t) = GM (t + b1 , t + b2 , · · · , t + bn ) − t (b) Prove that f is a monotonic increasing function, and that lim f (t) = AM (b1 , · · · , bn ) t→∞ 1 Original version is to show that sup Pn xi i=1 xi 2 +xi+1 xi+2 = n − 1. 49 M 41. (C1578, O. Johnson, C. S. Goodlad) For each fixed positive real number an , maximize a1 a2 · · · an (1 + a1 )(a1 + a2 )(a2 + a3 ) · · · (an−1 + an ) over all positive real numbers a1 , · · · , an−1 . M 42. (C1630, Isao Ashiba) Maximize a1 a2 + a3 a4 + · · · + a2n−1 a2n over all permutations a1 , · · · , a2n of the set {1, 2, · · · , 2n} M 43. (C1662, Murray S. Klamkin) Prove that x1 2r+1 x2 2r+1 xn 2r+1 4r r + + ··· ≥ (x1 x2 + x2 x3 + · · · + xn x1 ) s − x1 s − x2 s − xn (n − 1)n2r−1 where n > 3, r ≥ 21 , xi ≥ 0 for all i, and s = x1 + · · · + xn . Also, Find some values of n and r such that the inequality is sharp. M 44. (C1674, Murray S. Klamkin) Given positive real numbers r, s and an integer n > rs , find positive real numbers x1 , · · · , xn so as to minimize µ ¶ 1 1 1 + r + · · · + r (1 + x1 )s (1 + x2 )s · · · (1 + xn )s . x1 r x2 xn M 45. (C1691, Walther Janous) Let n ≥ 2. Determine the best upper bound of x1 x2 xn + + ··· + x2 x3 · · · xn + 1 x1 x3 · · · xn + 1 x1 x2 · · · xn−1 + 1 over all x1 , · · · , xn ∈ [0, 1]. M 46. (C1892, Marcin E. Kuczma) Let n ≥ 4 be an integer. Find the exact upper and lower bounds for the cyclic sum n X xi x + xi + xi+1 i=1 i−1 over all n-tuples of nonnegative numbers x1 , · · · , xn such that xi−1 + xi + xi+1 > 0 for all i. Of course, xn+1 = x1 , x0 = xn . Characterize all cases in which either one of these bounds is attained. M 47. (C1953, Murray S. Klamkin) Determine a necessary and sucient condition on real constants r1 , · · · , rn such that x1 2 + x2 2 + · + xn 2 ≥ (r1 x1 + r2 x2 + · · · + rn xn )2 holds for all real numbers x1 , · · · , xn . M 48. (C2018, Marcin E. Kuczma) How many permutations (x1 , · · · , xn ) of {1, 2, · · · , n} are there such that the cyclic sum |x1 − x2 | + |x2 − x3 | + · · · + |xn−1 − xn | + |xn − x1 | is (a) a minimum, (b) a maximum ? M 49. (C2214, Walther Janous) Let n ≥ 2 be a natural number. Show that there exists a constant C = C(n) such that for all x1 , · · · , xn ≥ 0 we have v u n n X uY √ xi ≤ t (xi + C) i=1 i=1 Determine the minimum C(n) for some values of n. (For example, C(2) = 1.) 50 M 50. (C2615, Murray S. Klamkin) Suppose that x1 , · · · , xn are non-negative numbers such that X xi 2 X (xi xi+1 )2 = n(n + 1) 2 where e the sums here and subsequently are symmetric over the subscripts q {1, · · · , n}. (a) Determine the P P maximum of xi . (b) Prove or disprove that the minimum of xi is n(n+1) . 2 M 51. (Turkey 1996) Given real numbers 0 = x1 < x2 < · · · < x2n , x2n+1 = 1 with xi+1 − xi ≤ h for 1 ≤ i ≤ n, show that n 1−h X 1+h < x2i (x2i+1 − x2i−1 ) < . 2 2 i=1 M 52. (Poland 2002) Prove that for every integer n ≥ 3 and every sequence of positive numbers x1 , · · · , xn at least one of the two inequalities is satsified : n X i=1 n xi n X xi n ≥ , ≥ . xi+1 + xi+2 2 x + x 2 i−2 i=1 i−1 Here, xn+1 = x1 , xn+2 = x2 , x0 = xn , x−1 = xn−1 . M 53. (China 1997) Let x1 , · · · , x1997 be real numbers satisfying the following conditions: √ √ 1 − √ ≤ x1 , · · · , x1997 ≤ 3, x1 + · · · + x1997 = −318 3 3 Determine the maximum value of x1 12 + · · · + x1997 12 . M 54. (C2673, George Baloglou) Let n > 1 be an integer. (a) Show that (1 + a1 · · · an )n ≥ a1 · · · an (1 + a1 n−2 ) · · · (1 + a1 n−2 ) for all a1 , · · · , an ∈ [1, ∞) if and only if n ≥ 4. (b) Show that 1 1 1 n + + ··· + ≥ a1 (1 + a2 n−2 ) a2 (1 + a3 n−2 ) an (1 + a1 n−2 ) 1 + a1 · · · an for all a1 , · · · , an > 0 if and only if n ≤ 3. (c) Show that 1 1 1 n + + ··· + ≥ n−2 n−2 n−2 a1 (1 + a1 ) a2 (1 + a2 ) an (1 + an ) 1 + a1 · · · an for all a1 , · · · , an > 0 if and only if n ≤ 8. M 55. (C2557, Gord Sinnamon,Hans Heinig) (a) Show that for all positive sequences {xi }  2 j n X k X n k X X X 1  . xi ≤ 2 xj  xk j=1 i=1 j=1 k=1 k=1 (b) Does the above inequality remain true without the factor 2? (c) What is the minimum constant c that can replace the factor 2 in the above inequality? M 56. (C1472, Walther Janous) For each integer n ≥ 2, Find the largest constant Cn such that Cn n X i=1 for all real numbers a1 , · · · , an satisfying X |ai | ≤ Pn i=1 1≤i
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