Find the number of real solutions to the system of equations $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$ My approach is naive:
Given
 $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+3$
What to do next?
Tried it using trigonometry by replacing $z^2$ by $tan^2\theta$ but could not get promising results.
Is there any trick to such genre of problems?

We have $x\ge0$, $y\ge0$ and $z\ge0$.
One possible solution is $x=y=z=0$.
If one of $x,y,z$ is non-zero, the other two are also non-zero.
Now suppose that all of them are non-zero.
$$xyz=\frac{8x^2y^2z^2}{(1+x^2)(1+y^2)(1+z^2)}\le\frac{8x^2y^2z^2}{(2x)(2y)(2z)}=xyz$$
The equality holds if and only if $x^2=y^2=z^2=1$ and this would imply that $x=y=z=1$.
There are totally two solutions.
