Example 8 Let $x, y, z, w$ be real numbers, not all zero, find the maximum value of $\frac{x y+2 y z+z w}{x^{2}+y^{2}+z^{2}+w^{2}}$. (1985 Austrian-Polish Joint Competition Problem)
Consider only the case where $x \geqslant 0, y \geqslant 0, z \geqslant 0, w \geqslant 0$.
Introduce positive constants $\alpha, \beta, \gamma$, then we have $\alpha^{2} x^{2}+y^{2} \geqslant 2 \alpha x y, \beta^{2} y^{2}+z^{2} \geqslant 2 \beta y z, \gamma^{2} z^{2}+$ $w^{2} \geqslant 2 \gamma z w$, which means $\frac{\alpha}{2} x^{2}+\frac{y^{2}}{2 \alpha} \geqslant x y, \beta y^{2}+\frac{z^{2}}{\beta} \geqslant 2 y z, \frac{\gamma z^{2}}{2}+\frac{w^{2}}{2 \gamma} \geqslant z w$.

Adding the above three inequalities, we get
$$\begin{array}{c}
\frac{\alpha}{2} x^{2}+\left(\frac{1}{2 \alpha}+\beta\right) y^{2}+\left(\frac{1}{\beta}+\frac{\gamma}{2}\right) z^{2}+\frac{w^{2}}{2 \gamma} \geqslant x y+2 y z+z w \\
\text { Let } \frac{\alpha}{2}=\frac{1}{2 \alpha}+\beta=\frac{1}{\beta}+\frac{\gamma}{2}=\frac{1}{2 \gamma} \text {, solving gives } \alpha=\sqrt{2}+1, \beta=1, \gamma=\sqrt{2}-1, \text { thus } \\
\frac{\sqrt{2}+1}{2}\left(x^{2}+y^{2}+z^{2}+w^{2}\right) \geqslant x y+2 y z+z w
\end{array}$$

Therefore,
$$\frac{x y+2 y z+z w}{x^{2}+y^{2}+z^{2}+w^{2}} \leqslant \frac{\sqrt{2}+1}{2}$$

That is, the maximum value of $\frac{x y+2 y z+z w}{x^{2}+y^{2}+z^{2}+w^{2}}$ is $\frac{\sqrt{2}+1}{2}$.