<s>[INST] 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?
 [/INST] 
Given $\;x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}\,$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+\color{red}{3}$

That's actually a good approach, and it does in fact work out nicely, but:


*

*in order to reverse the fractions, you have to assume $\,x,y,z \ne 0\,$; however $\,x=y=z=0\,$ is a solution, which you lose if you don't state that assumption upfront;

*you made a mistake in the calculations, the second line should rather be:
$$
\begin{align}
&\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2} \cdot \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\color{red}{\frac{3}{2}} \\[5px]
\iff\quad &\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} - 2 \cdot \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) + 3 = 0 \\[5px]
\iff\quad &\left(\frac{1}{x}-1\right)^2 + \left(\frac{1}{y}-1\right)^2 + \left(\frac{1}{z}-1\right)^2 = 0 \\[5px]
\iff\quad &x=y=z=1
\end{align}
$$
</s>