题目：若 $x \in (0, \frac{\pi}{2})$，则 $\frac{8}{\cos x} - 5\sin x$ 的最小值为
解：设 $\lambda > 0$
$$
\begin{aligned}
\frac{8}{\cos x} - 5\sin x
&= \frac{8}{\cos x} + \lambda\cos x - (\lambda\cos x + 5\sin x) \\
&\ge 4\sqrt{2\lambda} - \sqrt{\lambda^2+25}
\end{aligned}
$$
不等式等号成立的条件是
$$
\begin{cases}
\frac{8}{\cos x} = \lambda \cos x \\[0.5em]
\frac{\lambda}{\sqrt{\lambda^2 + 25}} = \cos x
\end{cases}
$$
可得 $\frac{\lambda^3}{\lambda^2+25} = 8 \Rightarrow \lambda^3 - 8\lambda^2 - 200 = 0$
$$
\Rightarrow (\lambda-10)(\lambda^2+2\lambda+20) = 0 \Rightarrow \lambda = 10
$$
则 $\cos x = \frac{2}{\sqrt{5}}$，$\sin x = \frac{1}{\sqrt{5}}$.
代入得 $\frac{8}{\cos x} - 5\sin x \ge 3\sqrt{5}$，
所以 $\frac{8}{\cos x} - 5\sin x$ 的最小值是 $3\sqrt{5}$