Example 5 Let non-negative numbers $\alpha, \beta, \gamma$ satisfy $\alpha+\beta+\gamma=\frac{\pi}{2}$, find the minimum value of the function
$$f(\alpha, \beta, \gamma)=\frac{\cos \alpha \cos \beta}{\cos \gamma}+\frac{\cos \beta \cos \gamma}{\cos \alpha}+\frac{\cos \gamma \cos \alpha}{\cos \beta}$$
Let's assume $\gamma \leqslant \alpha, \beta$, then $\gamma \in\left[0, \frac{\pi}{6}\right]$. Fix $\gamma$, since
$$\begin{aligned}
\frac{\cos \beta \cos \gamma}{\cos \alpha}+\frac{\cos \gamma \cos \alpha}{\cos \beta} & =2 \cos \gamma\left(\frac{\cos ^{2} \gamma}{\sin \gamma+\cos (\alpha-\beta)}+\sin \gamma\right) \\
\cos \alpha \cdot \cos \beta & =\frac{1}{2}(\sin \gamma+\cos (\alpha-\beta))
\end{aligned}$$

Then
$$\begin{aligned}
f & =\frac{1}{2} \frac{\sin \gamma+\cos (\alpha-\beta)}{\cos \gamma}+2 \cos \gamma\left(\sin \gamma+\frac{\cos ^{2} \gamma}{\sin \gamma+\cos (\alpha-\beta)}\right) \\
& =\sin 2 \gamma+\frac{1}{2} \frac{\sin \gamma+\cos (\alpha-\beta)}{\cos \gamma}+2 \cdot \frac{\cos ^{3} \gamma}{\sin \gamma+\cos (\alpha-\beta)}
\end{aligned}$$

Since $\sin \gamma+\cos (\alpha-\beta) \leqslant \sin \gamma+1 \leqslant \frac{3}{2} \leqslant 2 \cos ^{2} \gamma$, we have
$$\begin{array}{l}
f \geqslant \sin 2 \gamma+\frac{1}{2} \frac{1+\sin \gamma}{\cos \gamma}+2 \cdot \frac{\cos ^{3} \gamma}{1+\sin \gamma} \\
\quad=2 \cos \gamma+\frac{1}{2} \frac{1+\sin \gamma}{\cos \gamma} \\
=2 \cos \gamma+\frac{1}{2} \cot \left(\frac{\pi}{4}-\frac{\gamma}{2}\right)
\end{array}$$

Let $\theta=\frac{\pi}{4}-\frac{\gamma}{2} \in\left[\frac{\pi}{6}, \frac{\pi}{4}\right]$, then
$$\begin{aligned}
f & =2 \sin 2 \theta+\frac{1}{2} \cot \theta=\frac{4 \tan ^{2} \theta}{1+\tan ^{2} \theta}+\frac{1}{2} \cot \theta \\
& =\frac{5}{2}+\frac{1}{2} \frac{(1-\tan \theta)\left(5 \tan ^{2} \theta-4 \tan \theta+1\right)}{\tan \theta\left(1+\tan ^{2} \theta\right)} \\
& \geqslant \frac{5}{2}\left(\tan \theta \in\left[\frac{\sqrt{3}}{3}, 1\right]\right)
\end{aligned}$$

Therefore, when $\alpha=\beta=\frac{\pi}{4}, \gamma=0$, $f$ reaches its minimum value $\frac{5}{2}$.