Solve the following system of equations:

$$
\frac{2 x^{2}}{1+x^{2}}=y, \quad \frac{2 y^{2}}{1+y^{2}}=z, \quad \frac{2 z^{2}}{1+z^{2}}=x
$$
I. solution. If at least one of $x, y, z$ is zero, then from the equations it follows that all of them are, and $x=y=z=0$ is clearly a solution. Therefore, we can assume that none of $x, y, z$ is zero. Multiplying the three equations together, and then multiplying both sides of the resulting equation by $\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)$ (which is certainly not zero) and dividing by $x y z$ (which is also not zero), we get:

$$
8 x y z=\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)
$$

Since for any number $a$, $0 \leq(1-a)^{2}=1-2 a+a^{2}$, so $2 a \leq 1+a^{2}$, and equality holds only for $a=1$:

$$
2 x \cdot 2 y \cdot 2 z \leq\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)
$$

and equality holds only if $x=y=z=1$.

Therefore, the system of equations has two solutions: $x=y=z=0$ and $x=y=z=1$.

II. solution. The left side of the equations is non-negative, so we can assume that $x \geq 0 ; y \geq 0 ; z \geq 0$.

Since $(1-a)^{2}=1-2 a+a^{2} \geq 0,1+a^{2} \geq 2 a$, so if $a \geq 0$, then

$$
\frac{2 a^{2}}{1+a^{2}} \leq a
$$

and equality holds precisely when $a=0$ or $a=1$.

Using this, the following chain of inequalities holds:

$$
y=\frac{2 x^{2}}{1+x^{2}} \leq x=\frac{2 z^{2}}{1+z^{2}} \leq z=\frac{2 y^{2}}{1+y^{2}} \leq y
$$

This can only be true if equality holds everywhere. This, as we have seen, only holds if $x=y=z=0$ or $x=y=z=1$, which are indeed solutions to the equation.