Basics of olympiad inequalities by Riasat S.

By Riasat S.

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The above inequality is known as Schur’s inequality, after Issai Schur. Proof. Since the inequality is symmetric in a, b, c WLOG we may assume that a ≥ b ≥ c. Then the inequality is equivalent to (a − b)(an (a − c) − bn (b − c)) + cn (a − c)(b − c) ≥ 0, which is obviously true. 2 Jensen’s Inequality Suppose f is a convex function in [a, b]. Then the inequality f a1 + a2 + · · · + an n ≤ f (a1 ) + f (a2 ) + · · · + f (an ) n is true for all ai ∈ [a, b]. Similarly, if f is concave in the interval the sign of inequality turns over.

Applied restrictions with homogeneous expressions in the variables. For example, in order to show that a3 + b3 + c3 − 3abc ≥ 0, one may assume, WLOG, that abc = 1 or a + b + c = 1 etc. The reason is explained below. Suppose that abc = k 3 . Let a = ka , b = kb , c = kc . This implies a b c = 1, and our inequality becomes a 3 + b 3 + c 3 − 3a b c ≥ 0, which is the same as before. Therefore the restriction abc = 1 doesn’t change anything of the inequality. Similarly one might also assume a + b + c = 1.

Let a = ka , b = kb , c = kc . This implies a b c = 1, and our inequality becomes a 3 + b 3 + c 3 − 3a b c ≥ 0, which is the same as before. Therefore the restriction abc = 1 doesn’t change anything of the inequality. Similarly one might also assume a + b + c = 1. The reader is requested to find out how it works. 6 Homogenization This is the opposite of Normalization. It is often useful to substitute a = x/y, b = y/z, c = z/x, when the condition abc = 1 is given. Similarly when a + b + c = 1 we can substitute a = x/x + y + z, b = y/x + y + z, c = z/x + y + z to homogenize the inequality.

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