<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] It is easy to show that $$x,y,z>0$$ Clearly that $$x=y=z=0$$ is one solution. Now we assume $$x\geq y\geq z$$ this is equivalent to 
$$\frac{z^2}{1+z^2}\geq \frac{2x^2}{1+x^2}\geq \frac{2y^2}{1+y^2}$$ From here we get
$$\frac{z^2}{1+z^2}\geq \frac{x^2}{1+x^2}$$ this implies $$z\geq x$$ and so on, and we get
$$x=y=z$$ and $$1+x^2=2x$$ this means $$(x-1)^2=0$$ or $$x=y=z=1$$
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