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Here is an ingenious solution of (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) Second Solution. (based on work by an winter program participant) We obtain √ (a2 b + b 2 c + c 2 a) (ab 2 + bc 2 + ca 2 ) = 2√ 1 [b(a2 + bc) + c(b 2 + ca) + a(c 2 + ab)] [c(a 2 + bc) + a(b 2 + ca) + b(c 2 + ab)] ≥ 1 (√ bc(a 2 + bc) + √ ca(b 2 + ca) + √ ) ab(c 2 + ab) (Cauchy − Schwarz) 2 ≥ 3 2 √ 3 √bc(a2 + bc) · √ca(b 2 + ca) · √ab(c 2 + ab) (AM − GM) = 1 √ 3 (a3 + abc) (b 2 3 + abc) (c 3 + abc) + 3√ (a 3 + abc) (b 3 + abc) (c 3 + abc) ≥ 1 √ 3 2 √ a 2 3 · abc · 2 √ b 3 · abc · 2 √ c 3 · abc + 3√ (a 3 + abc) (b 3 + abc) (c 3 + abc) (AM − GM) = abc + 3√ (a 3 + abc) (b 3 + abc) (c 3 + abc). We now illustrate normalization techniques to establish classical theorems. Using the same idea in the proof of the Cauchy-Schwarz inequality, we find a natural generalization : Theorem 3.4.2. Let a ij (i, j = 1, · · · , n) be positive real numbers. Then, we have (a 11 n + · · · + a 1n n ) · · · (a n1 n + · · · + a nn n ) ≥ (a 11 a 21 · · · a n1 + · · · + a 1n a 2n · · · a nn ) n . Proof. Since the inequality is homogeneous, as in the proof of the theorem 11, we can normalize to (a i1 n + · · · + a in n ) 1 n = 1 or a i1 n + · · · + a in n = 1 (i = 1, · · · , n). Then, the inequality takes the form a 11 a 21 · · · a n1 + · · · + a 1n a 2n · · · a nn ≤ 1 or ∑ n i=1 a i1 · · · a in ≤ 1. Hence, it suffices to show that, for all i = 1, · · · , n, a i1 · · · a in ≤ 1 n , where a i1 n + · · · + a in n = 1. To finish the proof, it remains to show the following homogeneous inequality : Theorem 3.4.3. (AM-GM inequality) Let a 1 , · · · , a n be positive real numbers. Then, we have a 1 + · · · + a n n ≥ n√ a 1 · · · a n . Proof. Since it’s homogeneous, we may rescale a 1 , · · · , a n so that a 1 · · · a n = 1. 4 We want to show that a 1 · · · a n = 1 =⇒ a 1 + · · · + a n ≥ n. The proof is by induction on n. If n = 1, it’s trivial. If n = 2, then we get a 1 + a 2 − 2 = a 1 + a 2 − 2 √ a 1 a 2 = ( √ a 1 − √ a 2 ) 2 ≥ 0. Now, we assume that it holds for some positive integer n ≥ 2. And let a 1 , · · · , a n+1 be positive numbers such that a 1 · · · a n a n+1 =1. We may assume that a 1 ≥ 1 ≥ a 2 . (Why?) It follows that a 1 a 2 + 1 − a 1 − a 2 = (a 1 − 1)(a 2 − 1) ≤ 0 so that a 1 a 2 + 1 ≤ a 1 + a 2 . Since (a 1 a 2 )a 3 · · · a n = 1, by the induction hypothesis, we have a 1 a 2 + a 3 + · · · + a n+1 ≥ n. Hence, a 1 + a 2 − 1 + a 3 + · · · + a n+1 ≥ n. The following simple observation is not tricky : a i 4 Set x i = (a 1···a n ) n 1 (i = 1, · · · , n). Then, we get x 1 · · · x n = 1 and it becomes x 1 + · · · + x n ≥ n. 40
Let a, b > 0 and m, n ∈ N. Take x 1 = · · · = x m = a and x m+1 = · · · = x xm+n = b. Applying the AM-GM inequality to x 1 , · · · , x m+n > 0, we obtain ma + nb m + n ≥ (am b n ) 1 m+n or m m + n a + n m + n b ≥ a m n m+n b m+n . Hence, for all positive rationals ω 1 and ω 2 with ω 1 + ω 2 = 1, we get We immediately have ω 1 a + ω 2 b ≥ a ω 1 b ω 2 . Theorem 3.4.4. Let ω 1 , ω 2 > 0 with ω 1 + ω 2 = 1. For all x, y > 0, we have ω 1 x + ω 2 y ≥ x ω1 y ω2 . Proof. We can choose a positive rational sequence a 1 , a 2 , a 3 , · · · such that And letting b i = 1 − a i , we get From the previous observation, we have lim a n = ω 1 . n→∞ lim b n = ω 2 . n→∞ a n x + b n y ≥ x an y bn By taking the limits to both sides, we get the result. Modifying slightly the above arguments, we see that the AM-GM inequality implies that Theorem 3.4.5. (Weighted AM-GM inequality) Let ω 1 , · · · , ω n > 0 with ω 1 + · · · + ω n = 1. For all x 1 , · · · , x n > 0, we have ω 1 x 1 + · · · + ω n x n ≥ x 1 ω1 · · · x n ω n . Alternatively, we find that it is a straightforward consequence of the concavity of ln x. Indeed, the weighted Jensen’s inequality says that ln(ω 1 x 1 + · · · + ω n x n ) ≥ ω 1 ln(x 1 ) + · · · + ω n ln(x n ) = ln(x 1 ω1 · · · x n ω n ). Recall that the AM-GM inequality is used to deduce the theorem 18, which is a generalization of the Cauchy-Schwarz inequality. Since we now get the weighted version of the AM-GM inequality, we establish weighted version of the Cauchy-Schwarz inequality. Theorem 3.4.6. (Hölder) Let x ij (i = 1, · · · , m, j = 1, · · · n) be positive real numbers. Suppose that ω 1 , · · · , ω n are positive real numbers satisfying ω 1 + · · · + ω n = 1. Then, we have ( n∏ ∑ m ) ⎛ ⎞ ωj m∑ n∏ x ij ≥ ⎝ ω x j ij ⎠ . j=1 i=1 Proof. Because of the homogeneity of the inequality, as in the proof of the theorem 12, we may rescale x 1j , · · · , x mj so that x 1j + · · · + x mj = 1 for each j ∈ {1, · · · , n}. Then, we need to show that n∏ m∑ n∏ m∑ n∏ 1 ω j ω ≥ x j ω ij or 1 ≥ x j ij . j=1 i=1 j=1 The weighted AM-GM inequality provides that n∑ n∏ ω ω j x ij ≥ x j ij (i ∈ {1, · · · , m}) =⇒ j=1 j=1 However, we immediately have m∑ n∑ ω j x ij = i=1 j=1 n∑ j=1 i=1 m∑ ω j x ij = i=1 j=1 i=1 j=1 m∑ n∑ m∑ n∏ ω ω j x ij ≥ x j ij . i=1 j=1 ( n∑ m ) ∑ ω j x ij = j=1 i=1 i=1 j=1 n∑ ω j = 1. j=1 41
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