1. Determine all triples $(x, y, z)$ of real numbers for which

$$
x^{2}+y^{2}+z^{2} \leqq 6+\min \left\{x^{2}-\frac{8}{x^{4}}, y^{2}-\frac{8}{y^{4}}, z^{2}-\frac{8}{z^{4}}\right\}
$$

(J. Švrček)
Solution. If some triple $(x, y, z) \in \mathbb{R}^{3}(x y z \neq 0)$ satisfies the conditions of the problem, it is a solution to the following system of inequalities

$$
\begin{aligned}
x^{2}+y^{2}+z^{2} & \leqq 6+x^{2}-\frac{8}{x^{4}}, & & \frac{8}{x^{4}}+y^{2}+z^{2} \\
x^{2}+y^{2}+z^{2} & \leqq 6+y^{2}-\frac{8}{y^{4}}, & \text { i.e. } & x^{2}+\frac{8}{y^{4}}+z^{2} \leqq 6, \\
x^{2}+y^{2}+z^{2} & \leqq 6+z^{2}-\frac{8}{z^{4}}, & & x^{2}+y^{2}+\frac{8}{x^{4}} \leqq 6 .
\end{aligned}
$$

By adding all three inequalities of this system, we obtain the inequality

$$
\left(\frac{8}{x^{4}}+x^{2}+x^{2}\right)+\left(\frac{8}{y^{4}}+y^{2}+y^{2}\right)+\left(\frac{8}{z^{4}}+z^{2}+z^{2}\right) \leqq 18 .
$$

The expressions in each of the three parentheses on the left side can be estimated using the inequality between the arithmetic and geometric means of a triplet of positive numbers. We thus obtain successively

$$
\begin{aligned}
18 & \geqq\left(\frac{8}{x^{4}}+x^{2}+x^{2}\right)+\left(\frac{8}{y^{4}}+y^{2}+y^{2}\right)+\left(\frac{8}{z^{4}}+z^{2}+z^{2}\right) \geqq \\
& \geqq 3 \sqrt[3]{\frac{8}{x^{4}} \cdot x^{2} \cdot x^{2}}+3 \sqrt[3]{\frac{8}{y^{4}} \cdot y^{2} \cdot y^{2}}+3 \sqrt[3]{\frac{8}{z^{4}} \cdot z^{2} \cdot z^{2}}=18
\end{aligned}
$$

It follows that in each of the three used inequalities between the arithmetic and geometric means, equality occurs, so the corresponding triplet of numbers always has three identical components. Therefore, it must simultaneously hold that

$$
\frac{8}{x^{4}}=x^{2}, \quad \frac{8}{y^{4}}=y^{2}, \quad \frac{8}{z^{4}}=z^{2}
$$

i.e.

$$
x^{6}=y^{6}=z^{6}=8 \text {. }
$$

From the last condition, it immediately follows that

$$
(x, y, z)=\left(\varepsilon_{1} \sqrt{2}, \varepsilon_{2} \sqrt{2}, \varepsilon_{3} \sqrt{2}\right), \quad \text { where } \varepsilon_{i} \in\{-1 ; 1\} \text { for } i=1,2,3 \text {. }
$$

Given the used consequential transformation, it is necessary to perform a check, by which we find that all 8 triples of real numbers determined by relation (1) satisfy the conditions of the problem.