Example 9(1) Given the ellipse $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m>0, n>0)$ passes through the point $P(3 \sqrt{3}, 1)$, find the minimum value of $m+n$, and the equation of the ellipse when $m+n$ takes the minimum value.
Solution 1 Substitute the coordinates of point $P$ into the ellipse equation
$$
n=\frac{m}{\sqrt{m^{2}-27}},
$$

then
$$
\begin{aligned}
& (m+n)^{2}=m^{2}+\frac{2 m^{2}}{\sqrt{m^{2}-27}}+\frac{m^{2}}{m^{2}-27} \\
= & m^{2}+\frac{2\left(m^{2}-27\right)+54}{\sqrt{m^{2}-27}}+\frac{m^{2}-27+27}{m^{2}-27} \\
= & m^{2}+2 \sqrt{m^{2}-27}+\frac{27}{\sqrt{m^{2}-27}}+\frac{27}{\sqrt{m^{2}-27}}+\frac{27}{m^{2}-27}+1 \\
= & {\left[\left(m^{2}-27\right)+\frac{27}{\sqrt{m^{2}-27}}+\frac{27}{\sqrt{m^{2}-27}}\right]+\left[\sqrt{m^{2}-27}+\sqrt{m^{2}-27}+\frac{27}{m^{2}-27}\right]+28 } \\
\geqslant & 3 \sqrt[3]{27^{2}}+3 \sqrt[3]{27}+28=64,
\end{aligned}
$$

thus, $m+n \geqslant 8$.
At this point, $m^{2}-27=\frac{27}{\sqrt{m^{2}-27}}$ (or $\sqrt{m^{2}-27}=\frac{27}{m^{2}-27}$),
thus, $m=6$.
Then, $n=2$ is derived, hence the equation of the ellipse is
$$
\frac{x^{2}}{36}+\frac{y^{2}}{4}=1.
$$

Solution 2 Let $\frac{3 \sqrt{3}}{m}=\cos \alpha, \frac{1}{n}=\sin \alpha$,
then $m=\frac{3 \sqrt{3}}{\cos \alpha}, n=\frac{1}{\sin \alpha}$,
thus $(m+n)^{2}=\frac{27}{\cos ^{2} \alpha}+\frac{6 \sqrt{3}}{\sin \alpha \cos \alpha}+\frac{1}{\sin ^{2} \alpha}$
$$
\begin{array}{l}
=\frac{27\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)}{\cos ^{2} \alpha}+\frac{6 \sqrt{3}\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)}{\sin \alpha \cos \alpha}+\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{\sin ^{2} \alpha} \\
=\frac{27 \sin ^{2} \alpha}{\cos ^{2} \alpha}+\frac{6 \sqrt{3} \sin \alpha}{\cos \alpha}+\frac{6 \sqrt{3} \cos \alpha}{\sin \alpha}+\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}+28 \\
=\left(\frac{27 \sin ^{2} \alpha}{\cos ^{2} \alpha}+\frac{3 \sqrt{3} \cos \alpha}{\sin \alpha}+\frac{3 \sqrt{3} \cos \alpha}{\sin \alpha}\right)+\left(\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}+\frac{3 \sqrt{3} \sin \alpha}{\cos \alpha}+\frac{3 \sqrt{3} \sin \alpha}{\cos \alpha}\right)+28 \\
\geqslant 3 \sqrt[3]{27^{2}}+3 \sqrt[3]{27}+28=64 .
\end{array}
$$

thus, $m+n \geqslant 8$,
at this point $3 \sqrt{3} \sin ^{3} \alpha=\cos ^{3} \alpha$,
thus, $\tan \alpha=\frac{\sqrt{3}}{3}$.
Then, $m=6, n=2$ is derived. (Omitted below)
Solution 3 Since $\frac{27}{m^{2}}+\frac{1}{n^{2}}=1$,
then $\square$
$$
\begin{aligned}
& (m+n)^{2}=(m+n)^{2}\left(\frac{27}{m^{2}}+\frac{1}{n^{2}}\right) \\
= & \left(m^{2}+2 m n+n^{2}\right)\left(\frac{27}{m^{2}}+\frac{1}{n^{2}}\right) \\
= & 27+\frac{m^{2}}{n^{2}}+\frac{54 n}{m}+\frac{2 m}{n}+\frac{27 n^{2}}{m^{2}}+1 \\
= & \left(\frac{m^{2}}{n^{2}}+\frac{27 n}{m}+\frac{27 n}{m}\right)+\left(\frac{m}{n}+\frac{m}{n}+\frac{27 n^{2}}{m^{2}}\right)+28 \\
\geqslant & 3 \sqrt[3]{27^{2}}+3 \sqrt[3]{27}+28=64 .
\end{aligned}
$$

thus, $m+n \geqslant 8$, at this point $m^{3}=27 n^{3}$.
Thus, $m=3 n$,
then $m=6, n=2$ is solved. (Omitted below)