Olympiad Inequalities by Mildorf T.J.

By Mildorf T.J.

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42. ) Let n ≥ 2 be a positive integer and x1 , x2 , . . , xn , 2 y1 , y2 , . . , yn a sequence of 2n positive reals. Suppose z2 , z3 , . . , z2n is such that zi+j ≥ xi yj for all i, j ∈ {1, . . , n}. Let M = max{z2 , z3 , . . , z2n }. Prove that M + z2 + z3 + · · · + z2n 2n 2 ≥ x1 + · · · + xn n y1 + · · · + yn n Reid’s official solution. Let max(x1 , . . , xn ) = max(y1 , . √ . , yn ) = 1. ) We will prove M + z2 + · · · + z2n ≥ x1 + x2 + · · · + xn + y1 + y2 + · · · + yn , after which the desired follows by AM-GM.

USAMO 00/6) Let n ≥ 2 be an integer and S = {1, 2, . . , n}. Show that for all nonnegative reals a1 , a2 , . . , an , b1 , b2 , . . , bn , min{ai aj , bi bj } ≤ i,j∈S min{ai bj , aj bi } i,j∈S 29. (Kiran Kedlaya) Show that for all nonnegative a1 , a2 , . . , an , √ √ a1 + a1 a2 + · · · + n a1 · · · an a1 + a2 a1 + · · · + an ≤ n a1 · ··· n 2 n 30. (Vascile Cartoaje) Prove that for all positive reals a, b, c such that a + b + c = 3, a b c 3 + + ≥ ab + 1 bc + 1 ca + 1 2 31. (Gabriel Dospinescu) Prove that ∀a, b, c, x, y, z ∈ R+ | xy + yz + zx = 3, a(y + z) b(z + x) c(x + y) + + ≥3 b+c c+a a+b 32.

Hence, we made a logical choice. Inequality (3) is extremely sharp, and allowed us to obtain more a4 bc and a3 b3 terms simultaneously. In particular, it was necessary to cancel the a3 b3 terms. I’ll note that this inequality is peculiar to sixth degree symmetry in three variables - it does not belong to a family of similar, nice inequalities. Finally, inequality (4), which is a handy corollary to (3), is another Schur. Every inequality we have used so far is quite sharp, and so it is no surprise that the leftovers are the comparatively weak AM-GM.

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