
6. In the domain of real numbers solve the following 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 give an estimate of the LHS of the first equation. Consider the obvious inequality $4 x^{2} \leqslant x^{4}+4$ (it is equivalent to $0 \leqslant\left(x^{2}-2\right)^{2}$ ), which holds for any real number $x$ with the equality iff $x= \pm \sqrt{2}$. Then

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

Analogously we get next two inequalities. Thus we have

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

Summing up these inequalities we get

$$
x^{2}+y^{2}+z^{2} \leqslant x y+y z+z y,
$$

which is equivalent to

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

Thus

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

which are indeed solutions.

Conclusion. The solutions of the given system are the triples $(\sqrt{2}, \sqrt{2}, \sqrt{2})$ and $(-\sqrt{2},-\sqrt{2},-\sqrt{2})$.

