6. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& x^{4}+y^{2}+4=5 y z, \\
& y^{4}+z^{2}+4=5 z x, \\
& z^{4}+x^{2}+4=5 x y .
\end{aligned}
$$

(Jaroslav Švrček)
Solution. First, we estimate the left side of the first equation of the given system using the inequality $4 x^{2} \leqq x^{4}+4$, which is satisfied for any real number $x$, because it is equivalent to the inequality $0 \leqq\left(x^{2}-2\right)^{2}$. Equality in it occurs precisely when $x^{2}=2$, i.e., precisely when $x=\sqrt{2}$ or $x=-\sqrt{2}$.

We thus get

$$
4 x^{2}+y^{2} \leqq x^{4}+y^{2}+4=5 y z.
$$

Similarly, we estimate the left sides of the remaining two equations of the system. We thus obtain a triplet of inequalities

$$
4 x^{2}+y^{2} \leqq 5 y z, \quad 4 y^{2}+z^{2} \leqq 5 z x, \quad 4 z^{2}+x^{2} \leqq 5 x y
$$

whose summation gives us, after a simple rearrangement, the inequality

$$
x^{2}+y^{2}+z^{2} \leqq x y+y z+z y
$$

which we equivalently rearrange into the form

$$
(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \leqq 0 .
$$

The sum of squares cannot be negative, so equality must necessarily hold in inequality (2), hence $x=y=z$. Equality must therefore also hold in each inequality in (1). From this it follows that

$$
x=y=z=\sqrt{2} \quad \text { or } \quad x=y=z=-\sqrt{2} .
$$

By checking, we can easily verify that both found triples satisfy the given system.

Conclusion. The given system of equations has exactly two solutions in the domain of real numbers, namely the triples $(\sqrt{2}, \sqrt{2}, \sqrt{2})$ and $(-\sqrt{2},-\sqrt{2},-\sqrt{2})$.