1. ## Equality

What are the min and max values of:

$\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}$

for reals $x,y,z$ such that $xyz=1$?

EDIT: I now see that $1$ is the only possible value if only xyz=1, but I don't know how to reduce it to show that fact. Anyone?

2. $\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx} =$

$=\frac{1}{1+x+\frac{1}{z}}+\frac{1}{y+yz}+\frac{1} {1+z+zx}=$

$=\frac{z}{1+z+zx}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx }=$

$=\frac{1+z}{1+z+zx}+\frac{1}{1+y+yz}=$

$=\frac{1+z}{1+z+\frac{1}{y}}+\frac{1}{y+yz}=$

$=\frac{y+yz}{1+y+yz}+\frac{1}{1+y+yz}=$

$=\frac{1+y+yz}{1+y+yz}=1$

So, the expression is constant.