Bài Giảng Về Bất đẳng Thức

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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 presentstandard 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 Ho¨lder’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 paperOn 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 athojoolee@korea.com.

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 presentall 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!

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

a+ 1+ b² b+ 1 ≥ 1

3 I 2. (Columbia 2001) (x, y∈R)

3(x+y+ 1)²+ 1≥3xy I 3. (0< x, y <1)

x^y+y^x>1 I 4. (APMC 1993) (a, b≥0)

Ã√ a+√

b 2

!₂

≤ a+√³

a²b+√³ ab²+b

4 ≤a+√

ab+b

3 ≤

vu ut

Ã√3

a²+√³ b² 2

!₃

I 5. (Czech and Slovakia 2000) (a, b >0)

3

s

2(a+b) µ1

a+1 b

¶

≥³ ra

b +³ rb

a I 6. (Die√

W U RZEL, Heinz-J¨urgen Seiffert) (xy >0, x, y∈R) 2xy

x+y +

rx²+y²

2 ≥√

xy+x+y 2

I 7. (Crux Mathematicorum, Problem 2645, Hojoo Lee) (a, b, c >0) 2(a³+b³+c³)

abc + 9(a+b+c)² (a²+b²+c²) ≥33 I 8. (x, y, z >0)

√3

xyz+|x−y|+|y−z|+|z−x|

3 ≥x+y+z

3 I 9. (a, b, c, x, y, z >0)

p3

(a+x)(b+y)(c+z)≥√³

abc+√³ xyz

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I 10. (x, y, z >0)

x x+p

(x+y)(x+z)+ y y+p

(y+z)(y+x)+ z z+p

(z+x)(z+y)≤1 I 11. (x+y+z= 1, x, y, z >0)

√ x

1−x+ y

√1−y+ z

√1−z ≥ r3

2 I 12. (Iran 1998)

³1

x+¹_y+¹_z = 2, x, y, z >1

´

√x+y+z≥√

x−1 +p

y−1 +√ z−1 I 13. (KMO Winter Program Test 2001) (a, b, c >0)

p(a²b+b²c+c²a) (ab²+bc²+ca²)≥abc+p³

(a³+abc) (b³+abc) (c³+abc) I 14. (KMO Summer Program Test 2001) (a, b, c >0)

pa⁴+b⁴+c⁴+p

a²b²+b²c²+c²a²≥p

a³b+b³c+c³a+p

ab³+bc³+ca³ I 15. (Gazeta Matematic˜a, Hojoo Lee) (a, b, c >0)

pa⁴+a²b²+b⁴+p

b⁴+b²c²+c⁴+p

c⁴+c²a²+a⁴≥ap

2a²+bc+bp

2b²+ca+cp

2c²+ab I 16. (a, b, c∈R)

pa²+ (1−b)²+p

b²+ (1−c)²+p

c²+ (1−a)²≥3√ 2 2 I 17. (a, b, c >0) p

a²−ab+b²+p

b²−bc+c²≥p

a²+ac+c² I 18. (Belarus 2002) (a, b, c, d >0)

p(a+c)²+ (b+d)²+ 2|ad−bc|

p(a+c)²+ (b+d)² ≥p

a²+b²+p

c²+d²≥p

(a+c)²+ (b+d)² I 19. (Hong Kong 1998) (a, b, c≥1)

√a−1 +√

b−1 +√

c−1≤p

c(ab+ 1) I 20. (Carlson’s inequality) (a, b, c >0)

3

r(a+b)(b+c)(c+a)

8 ≥

rab+bc+ca 3 I 21. (Korea 1998) (x+y+z=xyz, x, y, z >0)

√ 1

1 +x² + 1

p1 +y² + 1

√1 +z² ≤3 2 I 22. (IMO 2001) (a, b, c >0)

√ a

a²+ 8bc+ b

√b²+ 8ca+ c

√c²+ 8ab ≥1 I 23. (IMO Short List 2004) (ab+bc+ca= 1, a, b, c >0)

3

r1

a+ 6b+ ³ r1

b + 6c+ ³ r1

c + 6a≤ 1 abc

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I 24. (a, b, c >0)

pab(a+b) +p

bc(b+c) +p

ca(c+a)≥p

4abc+ (a+b)(b+c)(c+a) I 25. (Macedonia 1995) (a, b, c >0)

r a b+c +

r b c+a+

r c a+b ≥2 I 26. (Nesbitt’s inequality) (a, b, c >0)

a

b+c + b

c+a+ c a+b ≥ 3

2 I 27. (IMO 2000) (abc= 1, a, b, c >0)

µ

a−1 +1 b

¶ µ

b−1 +1 c

¶ µ

c−1 +1 a

¶

≤1 I 28. ([ONI], Vasile Cirtoaje) (a, b, c >0)

µ a+1

b −1

¶ µ b+1

c −1

¶ +

µ b+1

c −1

¶ µ c+1

a−1

¶ +

µ c+1

a−1

¶ µ a+1

b −1

¶

≥3 I 29. (IMO Short List 1998) (xyz= 1, x, y, z >0)

x³

(1 +y)(1 +z)+ y³

(1 +z)(1 +x)+ z³

(1 +x)(1 +y) ≥ 3 4 I 30. (IMO Short List 1996) (abc= 1, a, b, c >0)

ab

a⁵+b⁵+ab + bc

b⁵+c⁵+bc+ ca

c⁵+a⁵+ca ≤1 I 31. (IMO 1995) (abc= 1, a, b, c >0)

1

a³(b+c)+ 1

b³(c+a)+ 1

c³(a+b) ≥ 3 2 I 32. (IMO Short List 1993) (a, b, c, d >0)

a

b+ 2c+ 3d+ b

c+ 2d+ 3a + c

d+ 2a+ 3b + d

a+ 2b+ 3c ≥2 3 I 33. (IMO Short List 1990) (ab+bc+cd+da= 1, a, b, c, d >0)

a³

b+c+d+ b³

c+d+a + c³

d+a+b + d³

a+b+c ≥ 1 3 I 34. (IMO 1968) (x1, x2>0, y1, y2, z1, z2∈R, x1y1> z12, x2y2> z22)

1

x1y1−z12 + 1

x2y2−z22 ≥ 8

(x1+x2)(y1+y2)−(z1+z2)² I 35. (Romania 1997) (a, b, c >0)

a²

a²+ 2bc+ b²

b²+ 2ca+ c²

c²+ 2ab ≥1≥ bc

a²+ 2bc+ ca

b²+ 2ca + ab c²+ 2ab I 36. (Canada 2002) (a, b, c >0)

a³ bc+ b³

ca +c³

ab ≥a+b+c

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I 37. (USA 1997) (a, b, c >0) 1

a³+b³+abc+ 1

b³+c³+abc+ 1

c³+a³+abc≤ 1 abc. I 38. (Japan 1997) (a, b, c >0)

(b+c−a)²

(b+c)²+a²+ (c+a−b)²

(c+a)²+b² + (a+b−c)² (a+b)²+c² ≥3

5 I 39. (USA 2003) (a, b, c >0)

(2a+b+c)²

2a²+ (b+c)² + (2b+c+a)²

2b²+ (c+a)² + (2c+a+b)² 2c²+ (a+b)² ≤8 I 40. (Crux Mathematicorum, Problem 2580, Hojoo Lee) (a, b, c >0)

1 a+1

b +1

c ≥ b+c

a²+bc+ c+a

b²+ca + a+b c²+ab I 41. (Crux Mathematicorum, Problem 2581, Hojoo Lee) (a, b, c >0)

a²+bc

b+c +b²+ca

c+a +c²+ab

a+b ≥a+b+c

I 42. (Crux Mathematicorum, Problem 2532, Hojoo Lee) (a²+b²+c²= 1, a, b, c >0) 1

a²+ 1 b²+ 1

c² ≥3 + 2(a³+b³+c³) abc I 43. (Belarus 1999) (a²+b²+c²= 3, a, b, c >0)

1

1 +ab+ 1

1 +bc+ 1 1 +ca ≥3

2

I 44. (Crux Mathematicorum, Problem 3032, Vasile Cirtoaje) (a²+b²+c²= 1, a, b, c >0) 1

1−ab+ 1

1−bc+ 1 1−ca ≤9

2 I 45. (Moldova 2005) (a⁴+b⁴+c⁴= 3, a, b, c >0)

1

4−ab+ 1

4−bc+ 1 4−ca ≤1 I 46. (Greece 2002) (a²+b²+c²= 1, a, b, c >0)

a

b²+ 1+ b

c²+ 1 + c a²+ 1 ≥ 3

4

³ a√

a+b√ b+c√

c´₂ I 47. (Iran 1996) (a, b, c >0)

(ab+bc+ca) µ 1

(a+b)² + 1

(b+c)² + 1 (c+a)²

¶

≥9 4 I 48. (Albania 2002) (a, b, c >0)

1 +√ 3 3√

3 (a²+b²+c²) µ1

a+1 b +1

c

¶

≥a+b+c+p

a²+b²+c² I 49. (Belarus 1997) (a, b, c >0)

a b +b

c +c

a ≥ a+b

c+a+b+c

a+b +c+a b+c

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I 50. (Belarus 1998, I. Gorodnin) (a, b, c >0) a

b +b c + c

a ≥a+b

b+c +b+c a+b+ 1 I 51. (Poland 1996)¡

a+b+c= 1, a, b, c≥ −³₄¢ a

a²+ 1+ b

b²+ 1 + c

c²+ 1 ≤ 9 10 I 52. (Bulgaria 1997) (abc= 1, a, b, c >0)

1

1 +a+b+ 1

1 +b+c + 1

1 +c+a ≤ 1

2 +a+ 1

2 +b+ 1 2 +c I 53. (Romania 1997) (xyz = 1, x, y, z >0)

x⁹+y⁹

x⁶+x³y³+y⁶ + y⁹+z⁹

y⁶+y³z³+z⁶ + z⁹+x⁹

z⁶+z³x³+x⁶ ≥2 I 54. (Vietnam 1991) (x≥y≥z >0)

x²y z +y²z

x +z²x

y ≥x²+y²+z² I 55. (Iran 1997) (x1x2x3x4= 1, x1, x2, x3, x4>0)

x³₁+x³₂+x³₃+x³₄≥max µ

x1+x2+x3+x4, 1 x1 + 1

x2 + 1 x3+ 1

x4

¶

I 56. (Hong Kong 2000) (abc= 1, a, b, c >0) 1 +ab²

c³ +1 +bc²

a³ +1 +ca²

b³ ≥ 18

a³+b³+c³ I 57. (Hong Kong 1997) (x, y, z >0)

3 +√ 3

9 ≥xyz(x+y+z+p

x²+y²+z²) (x²+y²+z²)(xy+yz+zx) I 58. (Czech-Slovak Match 1999) (a, b, c >0)

a

b+ 2c + b

c+ 2a+ c a+ 2b ≥1 I 59. (Moldova 1999) (a, b, c >0)

ab

c(c+a)+ bc

a(a+b)+ ca

b(b+c) ≥ a

c+a+ b

b+a+ c c+b I 60. (Baltic Way 1995) (a, b, c, d >0)

a+c

a+b+b+d

b+c +c+a

c+d+ d+b d+a ≥4 I 61. ([ONI], Vasile Cirtoaje) (a, b, c, d >0)

a−b

b+c +b−c

c+d+c−d

d+a+d−a a+b ≥0 I 62. (Poland 1993) (x, y, u, v >0)

xy+xv+uy+uv x+y+u+v ≥ xy

x+y + uv u+v

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I 63. (Belarus 1997) (a, x, y, z >0) a+y

a+xx+a+z

a+xy+a+x

a+yz≥x+y+z≥ a+z

a+zx+a+x

a+yy+a+y a+zz I 64. (Lithuania 1987) (x, y, z >0)

x³

x²+xy+y² + y³

y²+yz+z² + z³

z²+zx+x² ≥x+y+z 3 I 65. (Klamkin’s inequality) (−1< x, y, z <1)

1

(1−x)(1−y)(1−z)+ 1

(1 +x)(1 +y)(1 +z) ≥2 I 66. (xy+yz+zx= 1, x, y, z >0)

x

1 +x² + y

1 +y² + z

1 +z² ≥ 2x(1−x²)

(1 +x²)² +2y(1−y²)

(1 +y²)² +2z(1−z²) (1 +z²)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 1 + a

b

´ µ 1 +b

c

¶ ³ 1 + c

a

´

≥2 µ

1 +a+b+c

√3

abc

¶

I 69. (Elemente der Mathematik, Problem 1207, ˜Sefket Arslanagi´c) (x, y, z >0) x

y +y z +z

x≥ x+y+z

3√ 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−y²)²+ (1−z²)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+y³+z³+ 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≤ 1 27+176

27abcd 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)

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I 76. (Canada 1999) (x+y+z= 1, x, y, z≥0)

x²y+y²z+z²x≤ 4 27 I 77. (Hong Kong 1994) (xy+yz+zx= 1, x, y, z >0)

x(1−y²)(1−z²) +y(1−z²)(1−x²) +z(1−x²)(1−y²)≤ 4√ 3 9

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

3(ab+ac+ad+bc+bd+cd) I 79. (Poland 1998)¡

a+b+c+d+e+f = 1, ace+bdf ≥ ₁₀₈¹ 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≤a²b+b²c+c²a+ 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≥√ 3abc I 84. (Bearus 1996) (x+y+z=√

xyz, x, y, z >0)

xy+yz+zx≥9(x+y+z) I 85. (Poland 1991) (x²+y²+z²= 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 7

¡a⁴+b⁴+c⁴¢ I 89. (x, y, z≥0)

xyz≥(y+z−x)(z+x−y)(x+y−z) I 90. (Latvia 2002)³

1

1+a⁴ +_1+b¹4 +_1+c¹4 +_1+d¹4 = 1, a, b, c, d >0´ abcd≥3

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I 91. (Proposed for 1999 USAMO, [AB, pp.25]) (x, y, z >1)

x^x²^+2yzy^y²^+2zxz^z²^+2xy≥(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 abc≥4√

3 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+y²+z²≥√

2 (xy+yz) I 99. (APMC 1995) (m, n∈N, x, y >0)

(n−1)(m−1)(x^n+m+y^n+m) + (n+m−1)(xⁿy^m+x^myⁿ)≥nm(x^n+m−1y+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)

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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 andr denote the radii of the circumcircle and incircle of the triangleABC. Then, we haveR≥2r and the equality holds if and only ifABC is equilateral.

Proof. Let BC = a, CA = b, AB = c, s = â+b+c₂ and S = [ABC].² Recall the well-known identities : S = âbc_4R, S = rs, S² =s(s−a)(s−b)(s−c). Hence,R ≥ 2r is equivalent to âbc_4S ≥2^S_s or abc ≥8^S_s² 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 ifa=b=c.

First Proof. We use theRavi Substitution : Sincea,b,care the lengths of a triangle, there are positive reals x,y,zsuch thata=y+z,b=z+x,c=x+y. (Why?) Then, the inequality is (y+z)(z+x)(x+y)≥8xyz forx, 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 thata≥b≥c. It’s equivalent to

a³+b³+c³+ 3abc≥a²(b+c) +b²(c+a) +c²(a+b).

Sincec(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 realsa,b,c? Yes ! It’s possible to prove the inequality without the additional condition thata,b,c are the lengths of a triangle :

1The first geometric inequality is the Triangle Inequality : AB+BC≥AC

2In this book, [P] stands for the area of the polygonP.

3For example, we havec(a+b−c)−b(c+a−b) = (b−c)(b+c−a)≥0.

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Theorem 3. Letx,y,z >0. Then, we havexyz ≥(y+z−x)(z+x−y)(x+y−z). The equality holds if and only ifx=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 thaty+z ≤x. Then, xyz >0≥(y+z−x)(z+x−y)(x+y−z).

The inequality in the theorem 2 holds when some ofx,y,z are zeros :

Theorem 4. Let x,y,z≥0. Then, we havexyz ≥(y+z−x)(z+x−y)(x+y−z).

Proof. Sincex, y, z≥0, we can findpositive sequences{xn},{yn},{zn} for which

n→∞lim xn =x, lim

n→∞yn=y, lim

n→∞zn=z.

(For example, takexn =x+_n¹ (n= 1,2,· · ·), etc.) Applying the theorem 2 yields xnynzn ≥(yn+zn−xn)(zn+xn−yn)(xn+yn−zn) Now, taking the limits to both sides, we get the result.

Clearly, the equality holds whenx=y=z. However,xyz = (y+z−x)(z+x−y)(x+y−z) andx,y,z≥0 does not guarantee thatx=y=z. In fact, forx, y, z≥0, the equalityxyz= (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) Leta, b, c be positive numbers such thatabc= 1. Prove that µ

a−1 +1 b

¶ µ

b−1 +1 c

¶ µ

c−1 +1 a

¶

≤1.

First Solution. Sinceabc= 1, we make the substitution a= ^x_y,b = ^y_z, c = ^z_x forx, y,z > 0.⁵ We rewrite the given inequality in the terms ofx, y, z:

µx

y −1 +z y

¶ ³y

z −1 +x z

´ ³z

x−1 + y x

´

≤1 ⇔ xyz≥(y+z−x)(z+x−y)(x+y−z).

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) Leta,b,c be the lengths of the sides of a triangle. Prove that a²b(a−b) +b²c(b−c) +c²a(c−a)≥0.

Solution. After settinga=y+z,b=z+x,c=x+y forx, y, z >0, it becomes x³z+y³x+z³y≥x²yz+xy²z+xyz² or x²

y +y² z +z²

x ≥x+y+z, which follows from the Cauchy-Schwartz inequality

(y+z+x) µx²

y +y² z +z²

x

¶

≥(x+y+z)².

4See the theorem 10 in the chapter 3. Taker= 1.

5For example, takex= 1,y=¹_a,z=_ab¹.

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Problem 3. (IMO 1961/2, Weitzenb¨ock’s inequality)Let a,b,c be the lengths of a triangle with area S. Show that

a²+b²+c²≥4√ 3S.

Solution. Writea=y+z,b=z+x,c=x+y forx, 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 sidesa, b, c and area S, 2ab+ 2bc+ 2ca−(a²+b²+c²)≥4√

3S.

Exercise 3. (Pedoe’s inequality)Leta₁, b₁, c₁denote the sides of the triangleA₁B₁C₁ with areaF₁. Let a2, b2, c2 denote the sides of the triangleA2B2C2 with areaF2. Show that

a12(a22+b22−c22) +b12(b22+c22−a22) +c12(c22+a22−b22)≥16F1F2.

6Here, we used the well-known inequalitiesp²+q²≥2pqand (p+q+r)²≥3(pq+qr+rp).

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2.2 Trigonometric Substitutions

If you are faced with an integral that contains square root expressions such as Z p1−x²dx, Z p

1 +y²dy, Z p

z²−1dz

then trigonometric substitutions such as x= sint, y = tant, z = sect are very useful. When dealing with square root expressions, making a suitabletrigonometric substitution simplifies the given inequality.

Problem 4. (Latvia 2002)Let a,b,c,dbe the positive real numbers such that 1

1 +a⁴ + 1

1 +b⁴+ 1

1 +c⁴ + 1 1 +d⁴ = 1.

Prove thatabcd≥3.

Solution. We can writea² = tanA,b²= tanB, c² = tanC, d²= tanD, whereA, B, C, D∈¡ 0,^π₂¢

. Then, the algebraic identity becomes the following trigonometric identity :

cos²A+ cos²B+ cos²C+ cos²D= 1.

Applying the AM-GM inequality, we obtain

sin²A= 1−cos²A= cos²B+ cos²C+ cos²D≥3 (cosBcosCcosD)²³. Similarly, we obtain

sin²B ≥3 (cosCcosDcosA)²³,sin²C≥3 (cosDcosAcosB)²³, and sin²D≥3 (cosAcosBcosC)²³. Multiplying these inequalities, we get the result!

Exercise 4. ([ONI], Titu Andreescu, Gabriel Dosinescu)Let a,b,c,dbe 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 withx+y+z=xyz. Show that

√ 1

1 +x² + 1

p1 +y² + 1

√1 +z² ≤3 2.

Since the function f is not concave down on R⁺, we cannot apply Jensen’s inequality to the function f(t) = ^√_1+t¹ ₂. However, the functionf(tanθ) is concave down on ¡

0,^π₂¢

! Solution. We can write x = tanA, y = tanB, z = tanC, where A, B, C ∈ ¡

0,^π₂¢

. Using the fact that 1 + tan²θ=¡ ₁

cosθ

¢₂

, where cosθ6= 0, we rewrite it in the terms ofA, B,C : cosA+ cosB+ cosC≤ 3

2.

It follows from tan(π−C) =−z= _1−xy^x+y = tan(A+B) and fromπ−C, A+B∈(0, π) thatπ−C=A+B orA+B+C=π. Hence, it suffices to show the following.

Theorem 5. In any acute triangleABC, we havecosA+ cosB+ cosC≤ ³₂. Proof. Since cosxis concave down on¡

0,^π₂¢

, it’s a direct consequence of Jensen’s inequality.

We note that the function cosxis not concave down on (0, π). In fact, it’s concaveup on¡_π

2, π¢ . One may think that the inequality cosA+ cosB+ cosC≤ ³₂ doesn’t hold for any triangles. However, it’s known that it also holds for any triangles.

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Theorem 6. In any triangleABC, we havecosA+ cosB+ cosC≤ ³₂.

First Proof. It follows fromπ−C=A+B that cosC=−cos(A+B) =−cosAcosB+ sinAsinB or 3−2(cosA+ cosB+ cosC) = (sinA−sinB)²+ (cosA+ cosB−1)²≥0.

Second Proof. Let BC =a, CA= b, AB =c. Use the Cosine Law to rewrite the given inequality in the terms ofa,b,c :

b²+c²−a²

2bc +c²+a²−b²

2ca +a²+b²−c²

2ab ≤ 3

2. Clearing denominators, this becomes

3abc≥a(b²+c²−a²) +b(c²+a²−b²) +c(a²+b²−c²), which is equivalent toabc≥(b+c−a)(c+a−b)(a+b−c) in the theorem 2.

In case even when there is no condition such asx+y+z=xyz orxy+yz+zx= 1, the trigonometric substitutions are useful.

Problem 6. (APMO 2004/5)Prove that, for all positive real numbersa, b, c, (a²+ 2)(b²+ 2)(c²+ 2)≥9(ab+bc+ca).

Proof. ChooseA, B, C ∈â 0,^π₂đ

with a=√

2 tanA,b=√

2 tanB, andc=√

2 tanC. Using the well-known trigonometric identity 1 + tan²θ= _cos¹2θ, one may rewrite it as

4

9 ≥cosAcosBcosC(cosAsinBsinC+ sinAcosBsinC+ sinAsinBcosC). One may easily check the following trigonometric identity

cos(A+B+C) = cosAcosBcosC−cosAsinBsinC−sinAcosBsinC−sinAsinBcosC.

Then, the above trigonometric inequality takes the form 4

9 ≥cosAcosBcosC(cosAcosBcosC−cos(A+B+C)). Letθ= ^A+B+C₃ . Applying the AM-GM inequality and Jesen’s inequality, we have

cosAcosBcosC≤

ẾcosA+ cosB+ cosC 3

ả3

≤cos³θ.

We now need to show that

4

9 ≥cos³θ(cos³θ−cos 3θ).

Using the trigonometric identity

cos 3θ= 4 cos³θ−3 cosθ or cos 3θ−cos 3θ= 3 cosθ−3 cos³θ, it becomes

4

27 ≥cos⁴θâ

1−cos²θđ , which follows from the AM-GM inequality

Ếcos²θ

2 ·cos²θ 2 ·â

1−cos²θđả¹

3

≤ 1 3

Ếcos²θ

2 +cos²θ 2 +â

1−cos²θđả

=1 3.

One find that the equality holds if and only if tanA= tanB= tanC=^√¹₂ if and only ifa=b=c= 1.

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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

1−x² + y

1−y² + z

1−z² ≥3√ 3 2 .

Exercise 6. ([TZ], pp.127)Let x, y, z be positive real numbers such thatx+y+z=xyz. Prove that

√ x

1 +x² + y

p1 +y² + z

√1 +z² ≤3√ 3 2 .

Exercise 7. ([ONI], Florina Carlan, Marian Tetiva) Prove that if x, y, z > 0 satisfy the condition x+y+z=xyz then

xy+yz+zx≥3 +p

1 +x²+p

1 +y²+p 1 +z².

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)Leta andbbe positive real numbers. Prove that

√ 1

1 +a² + 1

√1 +b² ≥ 2

√1 +ab if either (1)0< a, b≤1or (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 thetrigonometric inequality cosA+ cosB+ cosC≤³₂. One may ask that

In any trianglesABC, is there anatural relation between cosA+ cosB+ cosC and ^R_r, whereR andr are the radii of the circumcircle and incircle ofABC ?

Theorem 7. Let R andr denote the radii of the circumcircle and incircle of the triangleABC. Then, we havecosA+ cosB+ cosC= 1 +_R^r.

Proof. Use the identitya(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 triangleABC with BC=a, CA=b,AB=c. Lets denote the semiperimeter ofABC. Verify the follwing identities⁷ :

(1) ab+bc+ca=s²+ 4Rr+r², (2) abc= 4Rrs,

(3) cosAcosB+ cosBcosC+ cosCcosA=^s²^−4R_4R²2^+r², (4) cosAcosBcosC= ^s²^−(2R+r)_4R2 ²

Exercise 12. (a) Letp, q, r be the positive real numbers such thatp²+q²+r²+ 2pqr= 1. Show that there exists an acute triangle ABC such that p= cosA,q= cosB,r= cosC.

(b) Let p, q, r ≥ 0 with p²+q²+r²+ 2pqr = 1. Show that there are A, B, C ∈ £ 0,^π₂¤

with p = cosA, q= cosB,r= cosC, andA+B+C=π.

7For more identities, see the exercise 10.

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Exercise 13. ([ONI], Marian Tetiva)Let x,y,z be positive real numbers satisfying the condition x²+y²+z²+ 2xyz = 1.

Prove that

(1) xyz≤¹₈,

(2) xy+yz+zx≤ ³₄, (3) x²+y²+z²≥ ³₄, and (4) xy+yz+zx≤2xyz+¹₂.

Exercise 14. ([ONI], Marian Tetiva)Let x,y,z be positive real numbers satisfying the condition x²+y²+z²=xyz.

Prove that

(1) xyz≥27,

(2) xy+yz+zx≥27, (3) x+y+z≥9, and

(4) 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 that0≤ab+bc+ca−abc≤2.

Solution. Notice thata, b, c >1 implies thata²+b²+c²+abc >4. Ifa≤1, then we haveab+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+q²+r²+ 2pqr= 1. By the exercise 12, we can write

a= 2 cosA, b= 2 cosB, c= 2 cosC for someA, B, C ∈h 0,π

2 i

withA+B+C=π.

We are required to prove

cosAcosB+ cosBcosC+ cosCcosA−2 cosAcosBcosC≤ 1 2. One may assume thatA≥ ^π₃ or 1−2 cosA≥0. Note that

cosAcosB+ cosBcosC+ cosCcosA−2 cosAcosBcosC= cosA(cosB+ cosC) + cosBcosC(1−2 cosA).

We apply Jensen’s inequality to deduce cosB+ cosC≤³₂−cosA. Note that 2 cosBcosC= cos(B−C) + cos(B+C)≤1−cosA. These imply that

cosA(cosB+ cosC) + cosBcosC(1−2 cosA)≤cosA µ3

2 −cosA

¶ +

µ1−cosA 2

¶

(1−2 cosA).

However, it’s easy to verify that cosA¡₃

2−cosA¢

+¡_1−cos_A

2

¢(1−2 cosA) = ¹₂. In the above solution, we showed that

cosAcosB+ cosBcosC+ cosCcosA−2 cosAcosBcosC≤ 1 2

holds for allacute triangles. Using the results (c) and (d) in the exercise (4), we can rewrite it in the terms ofR, 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) andB(x, y) are real quadratic formsαx²+βxy+γy² : ⁸

Exercise 15. Let R andr denote the radii of the circumcircle and incircle of the triangle ABC. Let s be the semiperimeter ofABC. Show that

16Rr−5r²≤s²≤4R²+ 4Rr+ 3r².

8For a proof, see [WJB].

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2.3 Algebraic Substitutions

We know that some inequalities in triangle geometry can be treated by theRavisubstitution andtrigonomet- ricsubstitutions. We can also transform the given inequalities into easier ones through some cleveralgebraic substitutions.

Problem 8. (IMO 2001/2) Leta,b,c be positive real numbers. Prove that

√ a

a²+ 8bc+ b

√b²+ 8ca+ c

√c²+ 8ab ≥1.

First Solution. To remove the square roots, we make the following substitution :

x= a

√a²+ 8bc, y= b

√b²+ 8ca, z= c

√c²+ 8ab. Clearly,x, y, z∈(0,1). Our aim is to show thatx+y+z≥1. We notice that

a²

8bc = x² 1−x², b²

8ac = y² 1−y², c²

8ab = z²

1−z² =⇒ 1 512 =

µ x² 1−x²

¶ µ y² 1−y²

¶ µ z² 1−z²

¶ . Hence, we need to show that

x+y+z≥1, where 0< x, y, z <1 and (1−x²)(1−y²)(1−z²) = 512(xyz)². However, 1> x+y+zimplies that, by the AM-GM inequality,

(1−x²)(1−y²)(1−z²)>((x+y+z)²−x²)((x+y+z)²−y²)((x+y+z)²−z²) = (x+x+y+z)(y+z) (x+y+y+z)(z+x)(x+y+z+z)(x+y)≥4(x²yz)¹⁴ ·2(yz)¹² ·4(y²zx)¹⁴ ·2(zx)¹² ·4(z²xy)¹⁴ ·2(xy)¹²

= 512(xyz)². This is a contradiction !

Problem 9. (IMO 1995/2) Leta, b, c be positive numbers such thatabc= 1. Prove that 1

a³(b+c)+ 1

b³(c+a)+ 1

c³(a+b)≥ 3 2.

First Solution. After the substitutiona=_x¹,b= ¹_y,c= ¹_z, we getxyz= 1. The inequality takes the form x²

y+z + y²

z+x+ z² x+y ≥3

2. It follows from the Cauchy-schwartz inequality that

[(y+z) + (z+x) + (x+y)]

µ x²

y+z + y²

z+x+ z² x+y

¶

≥(x+y+z)² so that, by the AM-GM inequality,

x²

y+z + y²

z+x+ z²

x+y ≥x+y+z

2 ≥3(xyz)¹³

2 =3

2.

We offer an alternative solution of the problem 5 :

(Korea 1998) Letx,y,z be the positive reals withx+y+z=xyz. Show that

√ 1

1 +x² + 1

p1 +y² + 1

√1 +z² ≤3 2.

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Second Solution. The starting point is lettinga=_x¹,b= ¹_y,c= ¹_z. We find thata+b+c=abcis equivalent to 1 =xy+yz+zx. The inequality becomes

√ x

x²+ 1 + y

py²+ 1 + z

√z²+ 1 ≤3 2

or x

px²+xy+yz+zx+ y

py²+xy+yz+zx+ z

pz²+xy+yz+zx ≤ 3 2

or x

p(x+y)(x+z)+ y

p(y+z)(y+x)+ z

p(z+x)(z+y)≤ 3 2. By the AM-GM inequality, we have

p x

(x+y)(x+z) =xp

(x+y)(x+z) (x+y)(x+z) ≤ 1

2

x[(x+y) + (x+z)]

(x+y)(x+z) = 1 2

µ x

x+z + x x+z

¶ . In a like manner, we obtain

p y

(y+z)(y+x) ≤ 1 2

µ y

y+z + y y+x

¶

and z

p(z+x)(z+y) ≤1 2

µ z

z+x+ z 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 numbersa, b, c, we have a

b+c + b

c+a+ c a+b ≥ 3

2.

Proof 1. After the substitution x=b+c, y=c+a, z=a+b, it becomes X

cyclic

y+z−x 2x ≥ 3

2 or X

cyclic

y+z x ≥6, which follows from the AM-GM inequality as following:

X

cyclic

y+z x = y

x+z x+z

y +x y +x

z +y z ≥6

µy x·z

x·z y ·x

y ·x z ·y

z

¶¹

6

= 6.

Proof 2. We make the substitution

x= a

b+c, y= b

c+a, z= c a+b. It follows that

X

cyclic

f(x) = X

cyclic

a

a+b+c = 1, where f(t) = t 1 +t. Since f is concave down on(0,∞), Jensen’s inequality shows that

f µ1

2

¶

= 2 3 = 1

3 X

cyclic

f(x)≥f

µx+y+z 3

¶ or f

µ1 2

¶

≥f

µx+y+z 3

¶ .

Since f is monotone decreasing, we have 1

2 ≤x+y+z

3 or X

cyclic

a

b+c =x+y+z≥ 3 2.

Bài Giảng Về Bất đẳng Thức

TOPICS IN INEQUALITIES

Hojoo Lee

Introduction

To Students

Recommended Reading List

Contents

Chapter 1

100 Problems

Chapter 2

Substitutions

2.1 Euler’s Theorem and the Ravi Substitution

2.2 Trigonometric Substitutions

2.3 Algebraic Substitutions