Example 7 Let $V_{1}$ and $V_{2}$ be the volumes of the circumscribed sphere and the inscribed sphere of a regular quadrilateral pyramid, respectively. Try to find the minimum value of $\frac{V_{1}}{V_{2}}$.
Solution: As shown in Figure 8, let the side length of the base of the regular quadrilateral pyramid be $a$, and the radii of the circumscribed sphere and the inscribed sphere be $R$ and $r$ respectively. Let $2 \theta$ be the angle between the side face and the base, then $r = \frac{a}{2} \tan \theta$. Let $PO_1 = h$, and the center of the circumscribed sphere be $O$. In the right triangle $\triangle OO_1C$, we have
$$
(h-R)^{2}+\left(\frac{\sqrt{2}}{2} a\right)^{2}=R^{2}.
$$

Thus, $R=\frac{h^{2}+\frac{a^{2}}{2}}{2 h}$.
Since $h=\frac{a}{2} \tan 2 \theta=\frac{a \tan \theta}{1-\tan ^{2} \theta}$, we have
$\frac{R}{r}=\frac{\tan ^{4} \theta+1}{2 \tan ^{2} \theta\left(1-\tan ^{2} \theta\right)}$.
Let $t=\tan ^{2} \theta, y=\frac{R}{r}$, then $y=\frac{t^{2}+1}{2 t(1-t)}$.
Thus, $(2 y+1) t^{2}-2 y t+1=0$.
By $\Delta \geqslant 0$ and $y>0$, we know $y \geqslant 1+\sqrt{2}$.
Therefore, when $\theta=\arctan \sqrt{\sqrt{2}-1}$, $y$ has the minimum value $\sqrt{2}+1$, and $\frac{V_{1}}{V_{2}}$ has the minimum value $7+5 \sqrt{2}$.