385*. Solve the system of equations:

$$
\left\{\begin{array}{l}
x=\frac{2 y^{2}}{1+z^{2}} \\
y=\frac{2 z^{2}}{1+x^{2}} \\
z=\frac{2 x^{2}}{1+y^{2}}
\end{array}\right.
$$
$\triangle$ From the system, it is clear that all unknowns $x, y$, and $z$ are non-negative.

Multiply all the equations of the system and reduce the resulting equation to an algebraic one:

$$
\begin{aligned}
& x y z=\frac{8 x^{2} y^{2} z^{2}}{\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)}, \quad x y z\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)=8 x^{2} y^{2} z^{2} \\
& x y z\left(\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)-8 x y z\right)=0
\end{aligned}
$$

We need to consider two cases.

1) Suppose $x y z=0$.

If any of the three unknowns is zero, then, as can be seen from the system, the other two unknowns are also zero. We get the solution $(0 ; 0 ; 0)$.

2) Suppose $x y z \neq 0$. Then

$$
\begin{gathered}
\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)-8 x y z=0 \\
8 x y z=\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right) \\
\frac{2 x}{1+x^{2}} \cdot \frac{2 y}{1+y^{2}} \cdot \frac{2 z}{1+z^{2}}=1
\end{gathered}
$$

Each fraction in the left part of the last equation is non-negative and does not exceed 1 (think about why), so the equality of their product to 1 is possible only when each of them is equal to 1:

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

From this, $x=y=z=1$.

Answer: $(0 ; 0 ; 0),(1 ; 1 ; 1)$.

Note that inequalities were applied in solving this problem.