8 Real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$, then the maximum value of $xy+yz$ is
$\qquad$ .
$8 \frac{\sqrt{2}}{2}$ Hint: Since
$$
\begin{aligned}
1 & =\left(x^{2}+\frac{1}{2} y^{2}\right)+\left(\frac{1}{2} y^{2}+z^{2}\right) \\
& \geqslant 2 \sqrt{\frac{1}{2} x^{2} y^{2}}+2 \sqrt{\frac{1}{2} y^{2} z^{2}} \\
& =\sqrt{2}(x y+y z),
\end{aligned}
$$

thus $x y+y z \leqslant \frac{\sqrt{2}}{2}$.
Furthermore, when $x=1, y=\frac{\sqrt{2}}{2}, z=1$, the equality is achieved, so the maximum value is $\frac{\sqrt{2}}{2}$.