If $x=y^2+z^2$, $y=z^2+x^2$, and $z=x^2+y^2$, prove $\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+1} = 1$ If $x=y^2+z^2$, $y=z^2+x^2$, and $z=x^2+y^2$,
prove
$$\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+1}  = 1$$
 A: First of all there is an evident solution $x=y=z=0$ for which the second identity is... false. 
We will assume $x,y,z \neq 0$ in the following.
A first remark is that, evidently $x,y,z >0$
Let us call (1),(2),(3) your equations.
Make the difference (1)-(2):
$$x-y=y^2-x^2 \ \iff x-y=-(x-y)(x+y)$$
Let us assume for a while that  $x-y=0$ : in this case, we would deduce that  $x+y=-1$ which is impossible due to the positivity of y o$x $ and $y$.
Therefore, $x=y$. 
For the same reason, using difference (1)-(3), we get $x=z$.
Replacing $y$ and $z$ by $x$ in any of the 3 equations, we get $x=2x^2$. As $x \ne 0$, we deduce that $x=\tfrac12$ and globally :
$$x=y=z=\tfrac12$$
We can check that with these values, the targetted relationship is indeed fullfilled.
A: Hint:
From the first two,
$$x-y=y^2-x^2$$ factors as $$(x-y)(x+y+1)=0$$
which defines a pair of planes, and similarly
$$(z-x)(z+x+1)=0.$$
Eliminate $y,z$ and plug in the last identity.

Note that
$$\frac{-(x+1)}{-(x+1)+1}=\frac{x+1}{x}$$
and the last identity has one of the forms
$$3t=1,\\2t+t^{-1}=1,\\t+2t^{-1}=1.$$
