<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] We only need to continue a little further on the path you took. The first term is 
$$\frac{1}{1+x+xy}.\tag{1}$$  
Consider the second term
$$\frac{1}{1+y+yz}.$$
Using your substitution, we (you) find that this is 
$$\frac{x}{1+x+xy}.\tag{2}$$
Consider the third term $\dfrac{1}{1+z+zx}$ and replace $z$ by $\frac{1}{xy}$, and $zx$ by $\frac{1}{y}$. We end up with
$$\frac{xy}{1+x+xy}.\tag{3}$$
Add  up the first term, the second, and the third, or equivalently (1), (2), and (3). We get
$$\frac{1}{1+x+xy}+\frac{x}{1+x+xy}+\frac{xy}{1+x+xy},$$
which is $1$.
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