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3.3 Normalizations In the previous sections, 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 √ a2 + 8bc + b √ b2 + 8ca + c √ c2 + 8ab ≥ 1. Third Solution. We make the substitution x = a a+b+c , y = b a+b+c , z = xf(x 2 + 8yz) + yf(y 2 + 8zx) + zf(z 2 + 8xy) ≥ 1, c a+b+c .2 The problem is where f(t) = 1 √ t . Since f is convex on R + and x + y + z = 1, we apply (the weighted) Jensen’s inequality to obtain xf(x 2 + 8yz) + yf(y 2 + 8zx) + zf(z 2 + 8xy) ≥ f(x(x 2 + 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(x 2 + 8yz) + y(y 2 + 8zx) + z(z 2 + 8xy). Using x + y + z = 1, we homogenize it as (x + y + z) 3 ≥ x(x 2 + 8yz) + y(y 2 + 8zx) + z(z 2 + 8xy). However, this is easily seen from (x + y + z) 3 − x(x 2 + 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. Fourth Solution. We make the substitution x = bc a , y = ca 2 b , z = ab 2 c . Then, we get xyz = 1 and the inequality 2 becomes 1 1 1 √ + √ + √ ≥ 1 1 + 8x 1 + 8y 1 + 8z which is equivalent to ∑ √ (1 + 8x)(1 + 8y) ≥ √ (1 + 8x)(1 + 8y)(1 + 8z). cyclic After squaring both sides, it’s equivalent to 8(x + y + z) + 2 √ (1 + 8x)(1 + 8y)(1 + 8z) ∑ √ 1 + 8x ≥ 510. Recall that xyz = 1. The AM-GM inequality gives us x + y + z ≥ 3, (1 + 8x)(1 + 8y)(1 + 8z) ≥ 9x 8 9 · 9y 8 9 · 9z 8 9 = 729 and ∑ Using these three inequalities, we get the result. cyclic cyclic √ 1 + 8x ≥ ∑ cyclic (IMO 1983/6) Let a, b, c be the lengths of the sides of a triangle. Prove that a 2 b(a − b) + b 2 c(b − c) + c 2 a(c − a) ≥ 0. 2 Dividing by a + b + c gives the equivalent inequalityPcyclic a r a+b+c a 2 (a+b+c) 2 + 8bc (a+b+c) 2 ≥ 1. √ 9x 8 9 ≥ 9(xyz) 4 27 = 9. 34
Second Solution. After setting a = y + z, b = z + x, c = x + y for x, y, z > 0, it becomes x 3 z + y 3 x + z 3 y ≥ x 2 yz + xy 2 z + xyz 2 or x2 y + y2 z + z2 x ≥ x + y + z. Since it’s homogeneous, we can restrict our attention to the case x + y + z = 1. Then, it becomes ( ) x ( y ) ( z yf + zf + xf ≥ 1, y z x) where f(t) = t 2 . Since f is convex on R, we apply (the weighted) Jensen’s inequality to obtain ( ) x ( y ) ( ( z yf + zf + xf ≥ f y · y z x) x y + z · y z + x · z ) = f(1) = 1. x Problem 26. (KMO Winter Program Test 2001) Prove that, for all a, b, c > 0, √ (a2 b + b 2 c + c 2 a) (ab 2 + bc 2 + ca 2 ) ≥ abc + 3√ (a 3 + abc) (b 3 + abc) (c 3 + abc) First Solution. Dividing by abc, it becomes √ (a c + b a + c ) ( √ (a c b a + a b c) + b ) ( ) ( ) 2 b ≥ abc + 3 bc + 1 2 c ca + 1 2 ab + 1 . After the substitution x = a b , y = b c , z = c a , we obtain the constraint xyz = 1. It takes the form √ √ (x ) ( (x + y + z) (xy + yz + zx) ≥ 1 + 3 y ) ( ) z + 1 x + 1 z y + 1 . From the constraint xyz = 1, we find two identities ( x ) ( y ) ( ) ( ) ( ) ( ) z + 1 x + 1 z x + z y + x z + y y + 1 = = (z + x)(x + y)(y + z), z x y (x + y + z) (xy + yz + zx) = (x + y)(y + z)(z + x) + xyz = (x + y)(y + z)(z + x) + 1. Letting p = 3√ (x + y)(y + z)(z + x), the inequality now becomes √ p 3 + 1 ≥ 1 + p. Applying the AM-GM inequality, we have p ≥ 3 √2 √ xy · 2 √ yz · 2 √ zx = 2. It follows that (p 3 +1)−(1+p) 2 = p(p+1)(p−2) ≥ 0. Problem 27. (IMO 1999/2) Let n be an integer with n ≥ 2. (a) Determine the least constant C such that the inequality ∑ 1≤i 0. Since the inequality is homogeneous, we may normalize to x 1 +· · ·+x n = 1. We denote F (x 1 , · · · , x n ) = ∑ x i x j (x 2 i + x 2 j). 3 I slightly modified his solution in [Au99]. 1≤i
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Page 3 and 4: Chapter 1 Geometric Inequalities It
Page 5 and 6: Exercise 3. (Darij Grinberg) Let a,
Page 7 and 8: Carlitz also observed that the Neub
Page 9 and 10: 1.2 Trigonometric Methods In this s
Page 11 and 12: Alternatively, one may obtain anoth
Page 13 and 14: 1.3 Applications of Complex Numbers
Page 15 and 16: which follows from the AM-GM inequa
Page 17 and 18: 2.2 Algebraic Substitutions We know
Page 19 and 20: Proof 3. As in the previous proof,
Page 21 and 22: Problem 16. (IMO Short-list 2001) L
Page 23 and 24: (IMO 2000/2) Let a, b, c be positiv
Page 25 and 26: Proof. We don’t employ the Ravi s
Page 27 and 28: Third Solution. (by an IMO 2005 con
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Page 33 and 34: Problem 23. (Tournament of Towns 19
Page 35: Second Solution. It’s equivalent
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Page 43 and 44: Let a, b > 0 and m, n ∈ N. Take x
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Page 49 and 50: 4.3 Majorization Inequality We say
Page 51 and 52: Lemma 4.4.1. Let f : (a, b) −→
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Page 55 and 56: M 24. (IMO Short List 1998) Let r 1
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Page 59 and 60: M 57. (China 2002) Given c ∈ ( 1
Page 61 and 62: P 12. Putnam 98B1 Find the minimum
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