8. Real numbers $x, y, z$ satisfy $x+y+z=0$, and $x^{2}+y^{2}+z^{2}=1$. Let $m$ be the maximum of $x^{2}, y^{2}, z^{2}$, then the minimum value of $m$ is $\qquad$ .
8. $\frac{1}{2}$.

Since substituting $-x, -y, -z$ for $x, y, z$ respectively does not change the conditions or the conclusion, by symmetry we can assume without loss of generality that $x \geqslant 0, y \geqslant 0, z \leqslant 0$.
Also, $z=-(x+y)$, so,
$2 z^{2}=z^{2}+(x+y)^{2}=x^{2}+y^{2}+z^{2}+2 x y \geqslant 1$.
Thus, $z^{2} \geqslant \frac{1}{2}$. Therefore, $m \geqslant z^{2} \geqslant \frac{1}{2}$.
When $x=\frac{\sqrt{2}}{2}, y=0, z=-\frac{\sqrt{2}}{2}$, $m=\frac{1}{2}$.
Hence, the minimum value of $m$ is $\frac{1}{2}$.