Authors: Agakhanov N.Kh., Bogdanov I.I.

Find all triples of real numbers $x, y, z$ such that $1+x^{4} \leq 2(y-z)^{2}, 1+y^{4} \leq 2(z-x)^{2}, 1+z^{4} \leq 2(x-y)^{2}$.
Let for definiteness $x \geq y \geq z$. Since $0 \leq\left(x^{2}-1\right)^{2}=\left(x^{4}+1\right)-2 x^{2}$, we have $2 x^{2} \leq 1+x^{4} \leq 2(y-z)^{2}$, from which $|x| \leq y-z$. Similarly,

$|z| \leq x-y$. Therefore, $|z|+|x| \leq(x-y)+(y-z)=x-z$. This is possible only if $x \geq 0, z \leq 0$, in which case the inequality becomes an equality. Thus, all intermediate inequalities also become equalities, that is, $2 x^{2}=1+x^{4}, 2 z^{2}=1+z^{4}$, from which $x^{2}=z^{2}=1$, that is, $x=1, z=-1$. Moreover, $|x|=y-z$, from which $y=0$. Verification shows that this answer is correct.

## Answer

$\{-1,0,1\}$.