Question

设x>y>z>0,若1/x-y+1/y-z+n/z-x>=0恒成立,则n的最大值是?

Answer 1

1/(x-y)+1/(y-z)+n/(z-x)>=0 等价于 1/(x-y)+1/(y-z)>=n/(x-z)

等价于 (x-z)*[1/(x-y)+1/(y-z)]>=n

设 a=x-y b=y-z a,b>0 则有(a+b)*(1/a+1/b)>=n 恒成立

左边=(a+b)*(1/a+1/b)=2+a/b+b/a>=4 等号成立当且仅当a=b>0 从而由恒成立必有n