PUMaC 2013 · 个人决赛（A 组） · 第 1 题

PUMaC 2013 — Individual Finals (Division A) — Problem 1

- ≤ + + 2 2 2 2 2 2 a + 2 b + 2 c + 2 6 ab + c 6 bc + a 6 ca + b 2 2 2 for any positive real numbers a , b and c satisfying a + b + c = 1.

解析

-  + + 2 2 2 2 2 2 a + 2 b + 2 c + 2 6 ab + c 6 bc + a 6 ca + b 2 2 2 for any positive real numbers a , b and c satisfying a + b + c = 1. Solution 2 2 2 2 2 2 By the AM-GM inequality, 3 a + 3 b 6 ab , 3 b + 3 c 6 bc and 3 c + 3 a 6 ca . Hence 1 1 1 1 1 1
-  + + . 2 2 2 2 2 2 2 2 2 2 2 2 3 a + 3 b + c 3 b + 3 c + a 3 c + 3 a + b 6 ab + c 6 bc + a 6 ca + b Now ✓ ◆ X X X 1 1 1 1 = 2 2 2 2 2 2 3 b + 3 c + a a + 2 3 2 a a + 2 cyc cyc cyc 2 X 3 a 1 = 2 2 (3 2 a )(2 + a ) cyc ✓ ◆ 2 2 2 2 X a b c a = 2 2 2 2 (3 2 a )(2 + a ) (3 2 a )(2 + a ) cyc ✓ ◆ 2 2 2 2 X a b a b = 2 2 2 2 (3 2 a )(2 + a ) (3 2 b )(2 + b ) cyc 2 2 2 X ( a b ) (1 + 2 a + 2 b ) = (3 2 a )(2 + a )(3 2 b )(2 + b ) cyc 0 from which the desired inequality follows. Remark: One may also prove the inequality X X X 1 1 1 = 2 2 2 2 2 2 2 3 b + 3 c + a a + 2 3 a + 2 b + 2 c cyc cyc cyc by without loss of generality assuming a b c , appealing to the majorization relation 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 (3 a + 3 b + c , 3 a + b + 3 c , a + 3 b + 3 c ) (3 a + 2 b + 2 c , 2 a + 3 b + 2 c , 2 a + 2 b + 3 c ) 1 and the convexity of the function f ( x ) = , and applying Karamata’s inequality. x