题目

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

提问时间：2020-10-23

答案

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)