10.5. Find all triples of real numbers $(x, y, z)$ that satisfy:
$$
\begin{array}{l}
1+x^{4} \leqslant 2(y-z)^{2}, \\
1+y^{4} \leqslant 2(z-x)^{2}, \\
1+z^{4} \leqslant 2(x-y)^{2} .
\end{array}
$$
10.5. The ternary real number groups that satisfy the conditions are
$$
\begin{array}{l}
(-1,0,1),(-1,1,0),(0,-1,1), \\
(0,1,-1),(1,-1,0),(1,0,-1) .
\end{array}
$$

By symmetry, without loss of generality, assume $x \geqslant y \geqslant z$. Then
$$
2 x^{2} \leqslant 1+x^{4} \leqslant 2(y-z)^{2} \text {, }
$$

which implies $|x| \leqslant y-z$.
Similarly, we get $|z| \leqslant x-y$.
Adding these two inequalities, we have
$$
|x|+|z| \leqslant(y-z)+(x-y)=x-z \text {. }
$$

Thus, $x \geqslant 0 \geqslant z$, and the inequality becomes an equality. This means that the intermediate inequalities all become equalities:
$$
2 x^{2}=1+x^{4}, 2 z^{2}=1+z^{4} \text {. }
$$

Hence, $x^{2}=z^{2}=1, x=1, z=-1$.
Substituting these into $|x|=y-z$, we get $y=0$.