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

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
$$3t=1,\\2t+t^{-1}=1,\\t+2t^{-1}=1.$$
