
1. Determine all triples $(x, y, z)$ of positive numbers satisfying the system of equations

$$
\begin{aligned}
& 2 x^{3}=2 y\left(x^{2}+1\right)-\left(z^{2}+1\right), \\
& 2 y^{4}=3 z\left(y^{2}+1\right)-2\left(x^{2}+1\right), \\
& 2 z^{5}=4 x\left(z^{2}+1\right)-3\left(y^{2}+1\right) .
\end{aligned}
$$


Solution. For any integer $k \geqslant 3$ and any $x \geqslant 0$ we have

$$
2 x^{k} \geqslant[(k-1) x-(k-2)]\left(x^{2}+1\right) .
$$

To see this, observe that, by AM-GM inequality,

$$
x^{k}+x^{k}+\underbrace{x+x+\ldots+x}_{(k-3) \text { times }} \geqslant(k-1) x^{3}
$$

and add it to

$$
(k-2)\left(x^{2}-2 x+1\right) \geqslant 0 \text {. }
$$

Note that we have equality if and only if $x=1$.

Therefore, for $x, y, z$ satisfying the system of equations, we have

$$
\begin{aligned}
2 y\left(x^{2}+1\right)-\left(z^{2}+1\right) & \geqslant(2 x-1)\left(x^{2}+1\right), \\
3 z\left(y^{2}+1\right)-2\left(x^{2}+1\right) & \geqslant(3 y-2)\left(y^{2}+1\right), \\
4 x\left(z^{2}+1\right)-3\left(y^{2}+1\right) & \geqslant(4 z-3)\left(z^{2}+1\right),
\end{aligned}
$$

or

$$
\begin{array}{r}
2(y-x)\left(x^{2}+1\right)+(x-z)(x+z) \geqslant 0, \\
3(z-y)\left(y^{2}+1\right)+2(y-x)(y+x) \geqslant 0, \\
4(x-z)\left(z^{2}+1\right)+3(z-y)(z+y) \geqslant 0 .
\end{array}
$$

Now suppose $x \geqslant \max \{y, z\}$. Then from the second inequality of (1) we infer that $y \leqslant z$ and

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

which, by the first inequality in (1), implies $x=y=z$.

If $y \geqslant \max \{x, z\}$, then $z \leqslant x$ by the third inequality in (1) and

$$
\begin{aligned}
3(z-y)\left(y^{2}+1\right)+2(y-x)(y+x) & \leqslant(x-y)\left(3\left(y^{2}+1\right)-2 y-2 x\right) \\
& \leqslant(x-y)\left(3 y^{2}-4 y+3\right) \leqslant 0,
\end{aligned}
$$

hence, by the second inequality in (1), $x=y=z$. ously,

Finally, if $z \geqslant \max \{x, y\}$, then $x \leqslant y$ by the first estimate in (1) and, as previ-

$$
4(x-z)\left(z^{2}+1\right)+3(z-y)(z+y) \leqslant(y-z)\left(4 z^{2}-6 z+4\right) \leqslant 0
$$

which again implies $x=y=z$. Thus we have equality in (0) and hence $x=y=z=1$. We easily check that this is the solution to the system.
