已知x,y,z∈R^+,x+y+z=xyz,且去1/(x+y)+1/(y+z)+1/(z+x)≤k恒成立,则k的取值范围是?

只须求出1/(x+y)+1/(y+z)+1/(z+x)的最大值即可知道k的范围.
∵x+y+z=xyz
∴1/(xy)+1/(yz)+1/(zx)=1
由柯西不等式知：
[1/(x+y)+1/(y+z)+1/(z+x)]^2
≤(1^2+1^2+1^2)[1/(x+y)^2+1/(y+z)^2+1/(z+x)^2]
而且(x+y)^2=x^2+y^2+2xy≥2xy+2xy=4xy
∴1/(x+y)^2≤1/(4xy)
同理：1/(y+z)^2≤1/(4yz),1/(z+x)^2≤1/(4zx)
∴1/(x+y)^2+1/(y+z)^2+1/(z+x)^2≤[1/(xy)+1/(yz)+1/(zx)]/4=1/4
∴[1/(x+y)+1/(y+z)+1/(z+x)]^2
≤(1^2+1^2+1^2)[1/(x+y)^2+1/(y+z)^2+1/(z+x)^2]
≤3/4
∴1/(x+y)+1/(y+z)+1/(z+x)≤√3/2
以上不等式等号成立的条件是x=y=z=√3
即1/(x+y)+1/(y+z)+1/(z+x)最大值为√3/2
∴当k≥√3/2时,1/(x+y)+1/(y+z)+1/(z+x)≤k恒成立
k∈(√3/2 ,+∞)