Example 1 Let $a, b, c, d$ be real numbers, not all zero, find
$$f=\frac{a b+2 b c+c d}{a^{2}+b^{2}+c^{2}+d^{2}}$$

the maximum value.
Assume the maximum value of $f$ is $M$, then
$$a b+2 b c+c d \leqslant M\left(a^{2}+b^{2}+c^{2}+d^{2}\right),$$

Therefore, to establish an inequality of the above form and find a set of values for $a, b, c, d$ that make the inequality an equality.

Let $\alpha, \beta, \gamma>0$, then
$$\frac{\alpha}{2} a^{2}+\frac{b^{2}}{2 \alpha} \geqslant a b, \beta b^{2}+\frac{c^{2}}{\beta} \geqslant 2 b c, \frac{\gamma}{2} c^{2}+\frac{d^{2}}{2 \gamma} \geqslant c d .$$

Adding the above three inequalities, we get
$$\begin{array}{c}
\frac{\alpha}{2} a^{2}+\left(\frac{1}{2 \alpha}+\beta\right) b^{2}+\left(\frac{1}{\beta}+\frac{\gamma}{2}\right) c^{2}+\frac{d^{2}}{2 \gamma} \\
\geqslant a b+2 b c+c d
\end{array}$$

Let $\frac{\alpha}{2}=\frac{1}{2 \alpha}+\beta=\frac{1}{\beta}+\frac{\gamma}{2}=\frac{1}{2 \gamma}$, we get $\gamma=\frac{1}{\alpha}, \beta=\frac{1}{2}\left(\alpha-\frac{1}{\alpha}\right)$, and $\alpha^{4}-$ $6 \alpha^{2}+1=0$, solving the equation gives $\alpha=\sqrt{2}+1$ (the other three solutions are not valid), thus
$$\frac{\sqrt{2}+1}{2}\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \geqslant a b+2 b c+c d$$

That is $\square$
$$f \leqslant \frac{\sqrt{2}+1}{2}$$

Furthermore, when $a=d=1, b=c=\sqrt{2}+1$, the inequality becomes an equality, hence the maximum value of $f$ is $\frac{\sqrt{2}+1}{2}$.