PUMaC 2014 · 代数（A 组） · 第 5 题

PUMaC 2014 — Algebra (Division A) — Problem 5

[ 5 ] Real numbers x, y, z satisfy the following equality: 2 2 2 4( x + y + z ) = x + y + z Let M be the maximum of xy + yz + zx , and let m be the minimum of xy + yz + zx . Find M + 10 m . ∞ ∑ 20 x x p n +1 3 n

解析

[ 5 ] Real numbers x, y, z satisfy the following equality: 2 2 2 4( x + y + z ) = x + y + z Let M be the maximum of xy + yz + zx , and let m be the minimum of xy + yz + zx . Find M + 10 m . Solution: 2 2 2 Let A = x + y + z, B = x + y + z , C = xy + yz + zx . The problem statement gives us 4 A = B . 1 2 2 Then, A = B + 2 C = 4 A + 2 C . So, C = ( A − 2) − 2. And by the inequality C ≤ B , it 2 2 follows that A = B + 2 C ≤ 3 B = 12 A. So, 0 ≤ A ≤ 12, which means that − 2 ≤ C ≤ 48. -2 √ √ is attained at (2 , − 2 , 2) and 48 is attained at (4, 4, 4). So, M + 10 m = 48 − 20 = 28 . ∞ ∑ 20 x x p n +1 3 n