[Example 2.4.2] For real numbers $x, y, z, w$ not all zero, if
$$
\frac{6 x y+5 y z+6 z w}{x^{2}+y^{2}+z^{2}+w^{2}} \leqslant f \text {, }
$$

find the minimum value of $f$.
In this problem, since the variables $x, y, z, w$ do not appear symmetrically in the numerator on the left side of the given inequality, directly applying the mean inequality cannot achieve the desired result.
Therefore, we generalize the inequality
$$
2 x y \leqslant x^{2}+y^{2}
$$
by introducing a parameter $\alpha$, transforming it into the following inequality:
$$
2 x y \leqslant \alpha x^{2}+\frac{1}{\alpha} y^{2} \quad (\alpha>0),
$$
Similarly, we can have
$$
\begin{array}{ll}
2 y z \leqslant \beta y^{2}+\frac{1}{\beta} z^{2} & (\beta>0), \\
2 z w \leqslant \gamma z^{2}+\frac{1}{\gamma} w^{2} & (\gamma>0).
\end{array}
$$
Using these three inequalities with parameters, we have
$$
\begin{aligned}
& \frac{6 x y+5 y z+6 z w}{x^{2}+y^{2}+z^{2}+w^{2}} \\
= & \frac{1}{2} \cdot \frac{6 \times 2 x y+5 \times 2 y z+6 \times 2 z w}{x^{2}+y^{2}+z^{2}+w^{2}} \\
\leqslant & \frac{1}{2} \cdot \frac{6\left(\alpha x^{2}+\frac{1}{\alpha} y^{2}\right)+5\left(\beta y^{2}+\frac{1}{\beta} z^{2}\right)+6\left(\gamma z^{2}+\frac{1}{\gamma} w^{2}\right)}{x^{2}+y^{2}+z^{2}+w^{2}} \\
= & \frac{1}{2} \cdot \frac{6 \alpha x^{2}+\left(5 \beta+\frac{6}{\alpha}\right) y^{2}+\left(6 \gamma+\frac{5}{\beta}\right) z^{2}+\frac{6}{\gamma} w^{2}}{x^{2}+y^{2}+z^{2}+w^{2}}.
\end{aligned}
$$
To make the variables $x, y, z, w$ disappear and leave only constants on the right side of the inequality, we need to satisfy the following equations:
$$
6 \alpha=5 \beta+\frac{6}{\alpha}=6 \gamma+\frac{5}{\beta}=\frac{6}{\gamma}.
$$
Thus,
$$
\alpha=\frac{1}{\gamma}.
$$
The above equations become
$$
\begin{array}{c}
5 \beta+6 \gamma=6 \gamma+\frac{5}{\beta}, \\
\beta^{2}=1, \beta=1.
\end{array}
$$
Furthermore,
$$
6 \alpha=\frac{6}{\alpha}+5,
$$
Solving this equation with $\alpha>0$ gives $\alpha=\frac{3}{2}$.
Therefore,
$$
\left\{\begin{array}{l}
\alpha=\frac{3}{2}, \\
\beta=1, \\
\gamma=\frac{2}{3}.
\end{array}\right.
$$
Thus,
$$
6 \alpha=\frac{6}{\alpha}+5 \beta=\frac{5}{\beta}+6 \gamma=\frac{6}{\gamma}=9.
$$
The above inequality becomes
$$
\begin{aligned}
\frac{6 x y+5 y z+6 z w}{x^{2}+y^{2}+z^{2}+w^{2}} & \leqslant \frac{1}{2} \cdot \frac{9\left(x^{2}+y^{2}+z^{2}+w^{2}\right)}{x^{2}+y^{2}+z^{2}+w^{2}} \\
& =\frac{9}{2}.
\end{aligned}
$$
The equality holds if and only if $x: y: z=1: \frac{3}{2}: \frac{3}{2}$, i.e., the minimum value of $f$ is $\frac{9}{2}$.

Introducing parameters to generalize inequalities can avoid the interference of specific data and focus on solving general problems.
For example, in this problem, we parameterize the inequality $x^{2}+y^{2} \geqslant 2 x y$ into
$$
\alpha x^{2}+\frac{1}{\alpha} y^{2} \geqslant 2 x y,
$$
which solves the problem.
This inequality can also be transformed into the following general form:
$$
x^{2}+\lambda^{2} y^{2} \geqslant 2 \lambda x y.
$$
If $y>0$, it can be further transformed into
$$
\frac{x^{2}}{y} \geqslant 2 \lambda x-\lambda^{2} y.
$$
The following problem is the 2nd problem of the 36th International Mathematical Olympiad for high school students. This problem has many solutions, but using the generalization strategy and introducing parameters for the proof is quite simple.