<s>[INST] Given that $xyz=1$ , find $\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}$? I think I solved this problem, but I don't feel $100$ percent sure of my solution. We have:
$xy=\large {\frac 1z}$
$xz=\large \frac 1y$
$yz=\large \frac 1x$
So let's substitute these into our sum:
$\large  \frac{1}{1+x+\frac 1z}+\frac{1}{1+y+\frac 1x}+\frac{1}{1+z+\frac 1y}$
If we rewrite with a common denominator we get
$\large \frac{z}{1+z+xz}+\frac {x}{1+x+xy}+\frac{y}{1+y+zy}$
If $\large \frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}=\frac {x}{1+x+xy} +\frac{y}{1+y+zy}+\frac{z}{1+z+xz}$ , $;$   then $(x, y, z)=1$ and we can compute that the sum is equal to $1$. 
The problem I have with this is that I found what $(x, y, z)$  $had$ to be, but I but I only had one weak restriction on the variables. Secondly, what are other way of doing this that I can learn from? Thanks. 
 [/INST] \begin{align*}
\frac{1}{1+x+xy}+\frac{1}{1+z+xz}+\frac{1}{1+y+yz}&=\frac{1}{1+x+xy}+\frac{xy}{xy(1+z+xz)}+\frac{x}{x(1+y+yz)}\\
&=\frac{1}{1+x+xy}+\frac{xy}{xy+1+x}+\frac{x}{x+xy+1}\\
&=1
\end{align*}
</s>