(15) Real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$, then the maximum value of $\sqrt{2} x y+y z$ is
$\qquad$ .
(15) $\frac{\sqrt{3}}{2}$ Hint:
$$
\begin{aligned}
1 & =x^{2}+y^{2}+z^{2} \\
& =\left(x^{2}+\frac{2}{3} y^{2}\right)+\left(\frac{1}{3} y^{2}+z^{2}\right) \\
& \geqslant 2 \sqrt{\frac{2}{3}} x y+2 \sqrt{\frac{1}{3}} y z \\
& =\frac{2}{\sqrt{3}}(\sqrt{2} x y+y z),
\end{aligned}
$$

From this, we can get $\sqrt{2} x y+y z \leqslant \frac{\sqrt{3}}{2}$, where the equality holds if and only if $x=\sqrt{\frac{1}{3}}, y=$
$$
\sqrt{\frac{1}{2}}, z=\sqrt{\frac{1}{6}} .
$$