<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] From the system of equations, you can infer that $x,y,z \ge 0$ and consider the function $f(a) = \dfrac{2a^2}{1+a^2}=2 - \dfrac{2}{1+a^2}$. Thus if $a > b>0 \implies a^2>b^2\implies 1+a^2>1+b^2\implies \dfrac{2}{1+a^2} < \dfrac{2}{1+b^2}\implies -\dfrac{2}{1+a^2} > -\dfrac{2}{1+b^2}\implies 2-\dfrac{2}{1+a^2} > 2-\dfrac{2}{a+b^2}\implies f(a) > f(b)\implies f$ is strictly increasing. Thus if we have $x > y > z \implies f(x) > f(y) > f(z)\implies y > z > x $, contradication to $x > z$. Thus at least two variables must be equal to each other, say $x = y \implies f(x) = f(y) \implies y = z \implies x = y =z\implies x = \dfrac{2x^2}{1+x^2}\implies x^3-2x^2+x = 0 \implies x(x^2-2x+1) = 0\implies x(x-1)^2 = 0 \implies x = 0, 1 \implies x = y = z =1$ or $x = y = z = 0$.
</s>