Example 1 Given that $\angle A, \angle B, \angle C$ are the interior angles of $\triangle ABC$. Find
$$
W=\operatorname{tg} \frac{A}{2} \operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2}+\operatorname{ctg} \frac{A}{2} \operatorname{ctg} \frac{B}{2} \operatorname{ctg} \frac{C}{2}
$$

the minimum value.
Solution: Given that $\angle A, \angle B, \angle C$ are the interior angles of a triangle, we know that $\operatorname{tg} \frac{A}{2}, \operatorname{tg} \frac{B}{2}, \operatorname{tg} \frac{C}{2}$ are all greater than 0. For simplicity, let $x, y, z$ represent $\operatorname{tg} \frac{A}{2}, \operatorname{tg} \frac{B}{2}, \operatorname{tg} \frac{C}{2}$, respectively. Then, from $x y + y z + z x = 1$, we can deduce that
$x y z = \sqrt{x y \cdot y z \cdot z x}$
$\leqslant \sqrt{\left(\frac{x y + y z + z x}{3}\right)^{3}} = \sqrt{\frac{1}{27}} = \frac{\sqrt{3}}{9} < 1$.
Furthermore, when $0 < a \leqslant b < 1$, $a + \frac{1}{a} \geqslant b + \frac{1}{b}$,
thus $x y z + \frac{1}{x y z} \geqslant \frac{\sqrt{3}}{9} + \frac{9}{\sqrt{3}} = \frac{28}{9} \sqrt{3}$,

which means that when $\angle A = \angle B = \angle C$, $W_{\text {min }} = \frac{28}{9} \sqrt{3}$.