$1+x^4\leq 2(y-z)^2$ and switching of $x,y,z$ Find all triples of real numbers $x,y,z$ such that $1+x^4\leq 2(y-z)^2$, $1+y^4\leq 2(z-x)^2$, and $1+z^4\leq 2(x-y)^2$.
Beside $(1,0,-1)$ and permutations, I can't find any others. We cannot have two equal numbers. Maybe the inequality $1+a^4\geq 2a^2$ helps? It yields $|x-y|\geq |z|$ and analogous.

The only solutions are $(-1,0,1)$ and permutations.
Subtracting $2x^2,$ $2y^2,$ and $2z^2$ respectively from each inequality gives
\begin{align*}
(1-x^2)^2&=1+x^4-2x^2\leq 2(y-z)^2-2x^2&=&-2(x-y+z)(x+y-z)\\
(1-y^2)^2&=1+y^4-2y^2\leq 2(z-x)^2-2y^2&=&-2(x+y-z)(-x+y+z)\\
(1-z^2)^2&=1+z^4-2z^2\leq 2(x-y)^2-2z^2&=&-2(-x+y+z)(x-y+z).
\end{align*}
Since the left-hand-sides are non-negative, these inequalities can be multiplied:
$$0\leq (1-x^2)^2(1-y^2)^2(1-z^2)^2\leq -8(-x+y+z)^2(x-y+z)^2(x+y-z)^2\leq 0.$$
So one of $-x+y+z,$ $x-y+z,$ or $x+y-z$ is zero. After a permutation of variables we have $x+y-z=0,$ which forces $(1-x^2)^2=(1-y^2)^2=0,$ so $x,y\in\{-1,1\}.$ We cannot have $x=y$ because $1\leq 1+z^4\leq 2(x-y)^2.$ So $x=-y\in\{-1,1\},$ which gives $z=x+y=0.$
