4. Determine the real values of $x, y, z$, for which

$$
\left\{\begin{array}{l}
\frac{2 x^{2}}{x^{2}+1}=y \\
\frac{3 y^{3}}{y^{4}+y^{2}+1}=z \\
\frac{4 z^{4}}{z^{6}+z^{4}+z^{2}+1}=x
\end{array}\right.
$$
## Solution.

Let $(x, y, z) \in \mathbb{R}^{3}$ be a solution of the system. We have $x, y, z \geqslant 0$. (1p)

From the inequality of means, we get

$$
\left\{\begin{array} { l } 
{ \frac { x ^ { 2 } + 1 } { 4 ^ { 2 } } \geqslant \sqrt { x ^ { 2 } \cdot 1 } = x } \\
{ \frac { y ^ { 2 } + y ^ { 2 } + 1 } { 3 } \geqslant \sqrt [ 3 ] { y ^ { 6 } } = y ^ { 2 } } \\
{ \frac { z ^ { 6 } + z ^ { 4 } + z ^ { 2 } + 1 } { 4 } \geqslant \sqrt [ 4 ] { z ^ { 1 2 } } = z ^ { 3 } }
\end{array} \quad \text { or } \left\{\begin{array}{l}
\frac{2 x^{2}}{x^{2}+1} \leq x \\
\frac{3 y^{3}}{y^{4}+y^{2}+1} \leq y \\
\frac{4 z^{4}}{z^{6}+z^{4}+z^{2}+1} \leq z
\end{array}\right.\right.
$$

It follows that $y \leq x, z \leq y$ and $x \leq z$, hence $x=y=z$. (1p)

We obtain the solutions $x_{1}=y_{1}=z_{1}=0$ and $x_{2}=y_{2}=z_{2}=1$. (2p)