Example 5 Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has an eccentricity of $\frac{\sqrt{6}}{3}$, and the distance from one endpoint of the minor axis to the right focus is $\sqrt{3}$.
(1) Find the equation of the ellipse $C$;
(2) Suppose the line $l$ intersects the ellipse $C$ at points $A$ and $B$, and the distance from the origin $O$ to $l$ is $\frac{\sqrt{3}}{2}$. Find the maximum value of the area of $\triangle A O B$.
Solution: (1) Let the semi-focal distance of the ellipse be $c$. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{c}{a}=\frac{\sqrt{6}}{3}, \\
a=\sqrt{3} .
\end{array}\right.
$$
Solving this, we get $c=\sqrt{2}, b=1$.
Therefore, the equation of the ellipse is $\frac{x^{2}}{3}+y^{2}=1$.
(2) Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$. Then the equation of the line $l$ is
$$
\begin{array}{l}
\left(y_{1}-y_{2}\right)\left(x-x_{2}\right)-\left(x_{1}-x_{2}\right)\left(y-y_{2}\right)=0 \\
\Leftrightarrow\left(y_{1}-y_{2}\right) x-\left(x_{1}-x_{2}\right) y+\left(x_{1} y_{2}-x_{2} y_{1}\right)=0 .
\end{array}
$$

Given that the distance from the origin $O$ to the line $l$ is $\frac{\sqrt{3}}{2}$, we have
$$
\frac{\left|x_{1} y_{2}-x_{2} y_{1}\right|}{\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}}=\frac{\sqrt{3}}{2} \text {. }
$$

Thus, $S_{\triangle A O B}=\frac{1}{2}|A B| \times \frac{\sqrt{3}}{2}$
$$
\begin{array}{l}
=\frac{1}{2}\left|x_{1} y_{2}-x_{2} y_{1}\right| \\
=\frac{\sqrt{3}}{2} \sqrt{\left(\frac{x_{1}}{\sqrt{3}} \cdot y_{2}+y_{1} \cdot \frac{-x_{2}}{\sqrt{3}}\right)^{2}} \\
\leqslant \frac{\sqrt{3}}{2} \sqrt{\left(\frac{x_{1}^{2}}{3}+y_{1}^{2}\right)\left(\frac{x_{2}^{2}}{3}+y_{2}^{2}\right)}=\frac{\sqrt{3}}{2} .
\end{array}
$$

The equality holds if and only if $\frac{x_{1}}{\sqrt{3}}=k y_{2}, y_{1}=-k \cdot \frac{x_{2}}{\sqrt{3}}$.

Taking $A(\sqrt{3}, 0) 、 B(0,1)\left(y_{1}=x_{2}=0\right)$, the area of $\triangle A O B$ reaches its maximum value $\frac{\sqrt{3}}{2}$.

Note: The key to applying the Cauchy-Schwarz inequality in this problem is to identify two sets of numbers $\frac{x_{1}}{\sqrt{3}} 、 y_{1}$ and $y_{2} 、-\frac{x_{2}}{\sqrt{3}}$, not only to ensure that $\frac{x_{1}^{2}}{3}+y_{1}^{2}=1, \frac{x_{2}^{2}}{3}+y_{2}^{2}=1$, which amplifies the area $S_{\triangle A O B}$ to a constant, but also to make the equality achievable.