5. Find the largest positive number $\lambda$, such that for any real numbers $x, y, z$ satisfying $x^{2}+y^{2}+z^{2}=1$, the inequality
$|\lambda x y+y z| \leqslant \frac{\sqrt{5}}{2}$ holds.
(Zhang Zhengjie)
5. Notice that
$$
\begin{array}{l}
1=x^{2}+y^{2}+z^{2}=x^{2}+\frac{\lambda^{2}}{1+\lambda^{2}} y^{2}+\frac{1}{1+\lambda^{2}} y^{2}+z^{2} \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(\lambda|x y|+|y z|) \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(|\lambda x y+y z|),
\end{array}
$$

and when $y=\frac{\sqrt{2}}{2}, x=\frac{\sqrt{2} \lambda}{2 \sqrt{\lambda^{2}+1}}, z=\frac{\sqrt{2}}{2 \sqrt{\lambda^{2}+1}}$, both equalities can be achieved simultaneously.
Therefore, $\frac{\sqrt{1+\lambda^{2}}}{2}$ is the maximum value of $|\lambda x y+y z|$.
Let $\frac{\sqrt{1+\lambda^{2}}}{2}=\frac{\sqrt{5}}{2}$, then $\lambda=2$.