Question

问关于不等式的几道题

1.已知x,y是正实数，且x+2y=1,求证xy≤1/8并指出等号成立条件
2.已知0<x<1，求当x取何值时，根号下x(1-x)最大？
要有步骤

Answer 1

1. x+2y=1, 两边乘以x:
2xy + x^2 = x
xy = (x - x^2)/2
xy - 1/8 = (x - x^2)/2 - 1/8
= -(4x^2 - 4x +1)/8
= -(2x -1)^2/8 ≤ 0
所以xy ≤ 1/8
x = 1/2时, -(2x -1)^2/8 = 0, xy = 1/8

2. sqrt[x(1 - x)] (sqrt: 平方根)
= sqrt(x - x^2)
= sqrt(-x^2 + x -1/4 + 1/4)
= sqrt [-(x - 1/2)^2 +1/4]
当x = 1/2时, (x - 1/2)^2 = 0, sqrt[x(1 - x)]最大 (=1/2)