若0〈x〈y〈m,等式1/(x的平方)+1/(y-x)的平方+1/(m-y)的平方≥9恒成立,求m的最大值
问题描述：
若0〈x〈y〈m,等式1/(x的平方)+1/(y-x)的平方+1/(m-y)的平方≥9恒成立,求m的最大值
答
(1+1+1)[1/x^2+1/(y-x)^2+1/(m-y)^2]>=[1/x+1/(y-x)+1/(m-y)]^2
=>[1/x^2+1/(y-x)^2+1/(m-y)^2]>=[1/x+1/(y-x)+1/(m-y)]^2/3
[(x+(y-x)+(m-y)][1/x+1/(y-x)+1/(m-y)]>={(x*1/x)^(1/2)+[(y-x)*1/(y-x)]^(1/2)+(m-y)*1/(m-y)]^(1/2)}^2=9 =>
[1/x+1/(y-x)+1/(m-y)]>=9/m
=>[1/x^2+1/(y-x)^2+1/(m-y)^2]>=[1/x+1/(y-x)+1/(m-y)]^2/3>=9/m^2
9/m^2>=9
=>m