HMMT 二月 2014 · 冲刺赛 · 第 16 题

HMMT February 2014 — Guts Round — Problem 16

[ 8 ] Suppose that x and y are positive real numbers such that x xy + 2 y = 8. Find the maximum 2 2 possible value of x + xy + 2 y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND Organization Team Team ID#

解析

[ 8 ] Suppose that x and y are positive real numbers such that x − xy + 2 y = 8. Find the maximum 2 2 possible value of x + xy + 2 y . √ √ u 72+32 2 2 2 Answer: Let u = x + 2 y . By AM-GM, u ≥ 8 xy , so xy ≤ √ . If we let xy = ku 7 8 1 where k ≤ √ , then we have 8 u (1 − k ) = 8 2 2 u (1 + k ) = x + xy + 2 y 1 + k that is, u (1 + k ) = 8 · . It is not hard to see that the maximum value of this expression occurs at 1 − k √ 1 √ 1 + 1 72 + 32 2 8 k = √ , so the maximum value is 8 · = . 1 √ 7 1 − 8 8