Example 20 Let $P, Q$ be any two points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, satisfying $\angle P O Q=\frac{\pi}{2}$ (here $O$ is the center of the ellipse), try to find the maximum and minimum values of $\frac{1}{|O P|}+\frac{1}{|O Q|}$.
To facilitate the calculation of $|OP|, |OQ|$, convert the coordinates of points $P, Q$ to polar coordinates.

Solution: Let $P(x_1, y_1), Q(x_2, y_2)$. Converting to polar coordinates, we can set $x_1 = \rho_1 \cos \theta, y_1 = \rho_1 \sin \theta, x_2 = \rho_2 \cos \left(\theta + \frac{\pi}{2}\right) = -\rho_2 \sin \theta, y_2 = \rho_2 \sin \left(\frac{\pi}{2} + \theta\right) = \rho_2 \cos \theta$. Substituting into the ellipse equation, we can find
$$
\begin{array}{c}
\rho_1 = \sqrt{\frac{a^2 b^2}{b^2 \cos^2 \theta + a^2 \sin^2 \theta}}, \rho_2 = \sqrt{\frac{a^2 b^2}{b^2 \sin^2 \theta + a^2 \cos^2 \theta}}. \\
\text{Thus, } \frac{1}{|OP|} + \frac{1}{|OQ|} = \frac{1}{\rho_1} + \frac{1}{\rho_2} = \frac{\sqrt{b^2 \cos^2 \theta + a^2 \sin^2 \theta} + \sqrt{b^2 \sin^2 \theta + a^2 \cos^2 \theta}}{ab}.
\end{array}
$$

Taking the Cartesian coordinates of points $M(a \cos \theta, b \sin \theta), N(a \sin \theta, b \cos \theta)$, we have
$$
\begin{aligned}
\frac{1}{|OP|} + \frac{1}{|OQ|} & = \frac{|ON| + |OM|}{ab} \leq \frac{|MN|}{ab} \\
& = \frac{\sqrt{a^2 (\cos \theta - \sin \theta)^2 + b^2 (\sin \theta - \cos \theta)^2}}{ab} \\
& = \frac{\sqrt{2(a^2 + b^2) \cos^2 \left(\frac{\pi}{4} + \theta\right)}}{ab} \leq \frac{\sqrt{2(a^2 + b^2)}}{ab}.
\end{aligned}
$$

The equality holds when $\theta = k\pi - \frac{\pi}{4} (k \in \mathbb{Z})$, so the maximum value of $\frac{1}{|OP|} + \frac{1}{|OQ|}$ is $\frac{\sqrt{2(a^2 + b^2)}}{ab}$.

To find the minimum value, we use the following inequality:
If $a \geq d \geq 0, c \geq b \geq 0$, then $\sqrt{a+b} + \sqrt{c+d} \geq \sqrt{a+c} + \sqrt{b+d}$, with equality if and only if $a = d$ or $b = c$. In fact,
$$
\begin{aligned}
& (\sqrt{a+b} + \sqrt{c+d}) - (\sqrt{a+c} + \sqrt{b+d}) \\
= & (\sqrt{a+b} - \sqrt{a+c}) + (\sqrt{c+d} - \sqrt{b+d}) \\
= & (b-c) \left[\frac{1}{\sqrt{a+b} + \sqrt{a+c}} + \frac{1}{\sqrt{c+d} + \sqrt{b+d}}\right] \geq 0,
\end{aligned}
$$

with equality if and only if $b = c$ or $a = d$.
Observing the expression for $\frac{1}{|OP|} + \frac{1}{|OQ|}$, note that $a^2 \sin^2 \theta \geq b^2 \sin^2 \theta \geq 0, a^2 \cos^2 \theta \geq b^2 \cos^2 \theta \geq 0$. By the above inequality, we get
$$
\frac{1}{|OP|} + \frac{1}{|OQ|} \geq \frac{\sqrt{a^2 \sin^2 \theta + a^2 \cos^2 \theta} + \sqrt{b^2 \cos^2 \theta + b^2 \sin^2 \theta}}{ab} = \frac{a + b}{ab}.
$$

The equality holds when $\theta = k\pi$ or $k\pi + \frac{\pi}{2}$ (where $k$ is an integer), so the minimum value of $\frac{1}{|OP|} + \frac{1}{|OQ|}$ is $\frac{a + b}{ab}$.