HMMT 二月 2010 · 冲刺赛 · 第 27 题

HMMT February 2010 — Guts Round — Problem 27

[ 15 ] Suppose that there are real numbers a, b, c ≥ 1 and that there are positive reals x, y, z such that x y z a + b + c = 4 x y z xa + yb + zc = 6 2 x 2 y 2 z x a + y b + z c = 9 . What is the maximum possible value of c ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

解析

[ 15 ] Suppose that there are real numbers a, b, c ≥ 1 and that there are positive reals x, y, z such that x y z a + b + c = 4 x y z xa + yb + zc = 6 2 x 2 y 2 z x a + y b + z c = 9 . What is the maximum possible value of c ? √ 3 Answer: 4 The Cauchy-Schwarz inequality states that given 2 sequences of n real numbers 2 2 2 2 2 2 2 x , x , . . . , x and y , y , . . . , y , then ( x + x + . . . + x )( y + y + . . . + y ) ≥ ( x y + x y + . . . + x y ) 1 2 n 1 2 n 1 1 2 2 n n 1 2 n 1 2 n x x x x/ 2 y/ 2 z/ 2 1 2 n with equality holding if and only if = = . . . = . Applying this to { a , b , c } and y y y 1 2 n x/ 2 y/ 2 z/ 2 x y z 2 x 2 y 2 z x y z 2 { xa , yb , zc } yields ( a + b + c )( x a + y b + z c ) ≥ ( xa + yb + zb ) with equality holding if and only if x = y = z . However, equality does hold (both sides evaluate to 36), so x = y = z . The second equation then 3 x x x 3 / 2 3 / 2 3 / 2 becomes x ( a + b + c ) = 6, which implies x = . Then we have a + b + c = 4. To maximize 2 √ 3 3 / 2 c , we minimize a and b by setting a = b = 1. Then c = 2 or c = 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND