The given conditions imply that $x,y,z>0.$ WLOG we may assume that $x=\max \{x,y,z\}$ and therefore $x\geq z,$ which means $$\frac{2z^2}{1+z^2}\geq z \Rightarrow \frac{2z}{1+z^2}\geq 1 \Rightarrow (z-1)^2\leq 0$$and we get that $z=1$ and therefore $x=1$ as well and $y=\frac{2}{1+1}=1.$ Thus we obtain that the system of equations has only one solution $(x,y,z)=(1,1,1).$

Solution: 1) We can suppose that $x,y,z>=0$, because the RHS is not negative. 2) We can see that $f(t)=\frac{2{{t}^{2}}}{1+{{t}^{2}}}$ it is monotonous increasing 3) We can suppose that $x\ge y\ge z$ 4) On the basis of the 2) and 3) result $\frac{2{{x}^{2}}}{1+{{x}^{2}}}\ge \frac{2{{y}^{2}}}{1+{{y}^{2}}}\ge \frac{2{{z}^{2}}}{1+{{z}^{2}}}\Leftrightarrow y\ge z\ge x$ so on the basis of the 2) result that we must have $x=y=z$. In this case we have $\frac{2{{x}^{2}}}{1+{{x}^{2}}}=x\Rightarrow x=y=z=1$

gobathegreat wrote: Find all real solutions of the equation: $$x=\frac{2z^2}{1+z^2}$$$$y=\frac{2x^2}{1+x^2}$$$$z=\frac{2y^2}{1+y^2}$$ Solution. From the fact that $\frac{2a^2}{1+a^2}\le a$ holds for any non-negative real number $a$, it follows that \begin{align*}x=\frac{2z^2}{1+z^2}\wedge z=\frac{2y^2}{1+y^2}\wedge y=\frac{2x^2}{1+x^2} \Longrightarrow x\le z\le y\le x\Longrightarrow x=y=z, \end{align*}which results in $x=y=z=1$ or $0$ by the original equations. $\blacksquare$

And if we change the problem by asking to solve in complex numbers... $(1,1,1)$ (double root), $(0,0,0)$, and cyclic permutations of $(A,B,C), (\bar{A},\bar{B},\bar{C})$. $A,B,C=\frac{1}{267} \left(-25+34 i \sqrt{2}+e^{m\pi i/3}\sqrt[3]{565484+12376 i \sqrt{2}-75828 i \sqrt{3}-194376 \sqrt{6}}+e^{n\pi i/3}\sqrt[3]{565484+12376 i \sqrt{2}+75828 i \sqrt{3}+194376 \sqrt{6}}\right)$ Key: $A:(m=2,n=0)$, $B:(m=0,n=2)$, $C:(m=n=4)$

Note that $x,y,z\geq0$ if any number is 0, the other numbers are also 0 so $(x,y,z)=(0,0,0)$. Now if $x,y,z>0$ note that the system is analogous to: $\frac{1}{x}=\frac{1}{2z^2}+\frac{1}{2}$ $\frac{1}{y}=\frac{1}{2x^2}+\frac{1}{2}$ $\frac{1}{z}=\frac{1}{2y^2}+\frac{1}{2}$ Let $\frac{1}{x}=a$, $\frac{1}{y}=b$, $\frac{1}{z}=c$, then: $2a=c^2+1$ $2b=a^2+1$ $2c=b^2+1$ Adding all equation we obtain $(a-1)^2+(b-1)^2+(c-1)^2=0$ then $a=b=c=1$ so $(x,y,z)=(1,1,1)$ and are all solutions