Example 2 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 holds:
$$|\lambda x y+y z| \leqslant \frac{\sqrt{5}}{2} .$$
Since
$$\begin{aligned}
1 & =x^{2}+y^{2}+z^{2}=x^{2}+\frac{\lambda^{2}}{1+\lambda^{2}} y^{2}+\frac{1}{1+\lambda^{2}} y^{2}+z^{2} \\
& \geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(\lambda|x y|+|y z|) \geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(|\lambda x y+y z|),
\end{aligned}$$

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 $\frac{\sqrt{1+\lambda^{2}}}{2} \leqslant \frac{\sqrt{5}}{2}$, solving for $\lambda$ gives $|\lambda| \leqslant 2$. Hence, the maximum value of $\lambda$ is 2.