设x>0,y>0,且x눀+y눀⼀2=1,求x乘以根号下(1+y눀)的最小值

2025-05-14 07:24:08
推荐回答(1个)
回答1:

应该是最大值吧
y^2=2(1-x^2)
[x√(1+y^2)]^2=x^2(1+y^2)
=2x^2(3/2 - x^2)
<=2[(x^2+(3/2-x^2))/2]^2=9/8,
当x^2=3/4时等号可以取到
所以x√(1+y^2)最大值为3√2/4