PUMaC 2014 · 代数（A 组） · 第 7 题

PUMaC 2014 — Algebra (Division A) — Problem 7

[ 7 ] x, y, z are positive real numbers that satisfy x +2 y +6 z = 1. Let k be the maximum possible n value of 2 x + y + 3 z . Let n be the smallest positive integer such that k is an integer. Find the n value of k + n .

解析

[ 7 ] x, y, z are positive real numbers that satisfy x + 2 y + 6 z = 1. Let k be the maximum n possible value of 2 x + y + 3 z . Let n be the smallest positive integer such that k is an integer. n Find the value of k + n . Solution: − 1 / 3 1 / 3 2 / 3 − 1 / 3 1 / 3 We see that by holder inequality, we have 2 x + y +3 z = (2)( x )+(2 )(2 y )+(3 2 )(6 z ) ≤ 2 / 3 3 / 2 3 / 2 3 3 3 1 / 3 3 / 2 − 1 / 3 2 / 3 − 1 / 3 5 / 3 3 ( x + 2 y + 6 z ) (2 + 2 + (3 2 ) ) = 2 . Hence n = 3 and k + 3 = 35