8. Given real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$. Then the maximum value of $x y+y z$ is $\qquad$ .
8. $\frac{\sqrt{2}}{2}$.
$$
\begin{array}{l}
\text { From }\left(x^{2}+\frac{1}{2} y^{2}\right)+\left(\frac{1}{2} y^{2}+z^{2}\right) \\
\geqslant 2 \sqrt{\frac{1}{2} x^{2} y^{2}}+2 \sqrt{\frac{1}{2} y^{2} z^{2}} \\
=\sqrt{2}(x y+y z),
\end{array}
$$

we get $x y+y z \leqslant \frac{\sqrt{2}}{2}$.
Therefore, the maximum value is $\frac{\sqrt{2}}{2}$.