Question 11 Given $\alpha+\beta+\gamma=180^{\circ}(\alpha, \beta, \gamma \geqslant$ $0)$. Find the maximum value of $3 \cos \alpha+4 \cos \beta+5 \cos \gamma$.
Solve: In inequality (1), let
$$
2 y z=3,2 z x=4,2 x y=5 \text {, }
$$

i.e., $x=\sqrt{\frac{4 \times 5}{2 \times 3}}, y=\sqrt{\frac{3 \times 5}{2 \times 4}}, z=\sqrt{\frac{3 \times 4}{2 \times 5}}$,
and $\alpha=\angle A, \beta=\angle B, \gamma=\angle C$. Then
$3 \cos \alpha+4 \cos \beta+5 \cos \gamma \leqslant \frac{769}{120}$.
Equality holds if and only if $\sin \alpha: \sin \beta: \sin \gamma=\frac{1}{3}: \frac{1}{4}: \frac{1}{5}$.

By the Law of Sines, to determine $\alpha, \beta, \gamma$, it is sufficient to construct a triangle with side lengths $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$. Therefore, the maximum value sought is $\frac{769}{120}$.