Four. (20 points) Given the ellipse $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m>n$ $>0)$ passes through the fixed point $P(a, b)(a, b$ are real numbers, and $a b \neq$ $0,|a| \neq|b|)$. Find the minimum value of $m+n$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
From the given, we have $\frac{a^{2}}{m^{2}}+\frac{b^{2}}{n^{2}}=1$, which means
$$
\left(\frac{a}{m}\right)^{2}+\left(\frac{b}{n}\right)^{2}=1 \text {. }
$$

Given $a b \neq 0$, by symmetry, we can assume $a, b \in \mathbf{R}_{+}$.
From equation (1), let $\frac{a}{m}=\cos \alpha, \frac{b}{n}=\sin \alpha$ (where $\alpha$ is an acute angle).
Then we have $m=\frac{a}{\cos \alpha}, n=\frac{b}{\sin \alpha}$.
Thus, $(m+n)^{2}=\left(\frac{a}{\cos \alpha}+\frac{b}{\sin \alpha}\right)^{2}$
$$
\begin{array}{l}
=a^{2} \sec ^{2} \alpha+b \csc ^{2} \alpha+\frac{2 a b}{\sin \alpha \cdot \cos \alpha} \\
=a^{2}\left(\tan ^{2} \alpha+1\right)+b\left(\cot ^{2} \alpha+1\right)+\frac{2 a b\left(1+\tan ^{2} \alpha\right)}{\tan \alpha} .
\end{array}
$$

Let $t=\tan \alpha(t>0)$, then we have
$$
\begin{array}{l}
(m+n)^{2}=a^{2} t^{2}+\frac{b^{2}}{t^{2}}+2 a b t+\frac{2 a b}{t}+a^{2}+b^{2} \\
=a\left(a t^{2}+\frac{b}{t}+\frac{b}{t}\right)+b\left(\frac{b}{t^{2}}+a t+a t\right)+a^{2}+b^{2} \\
\geqslant a \cdot 3 \sqrt[3]{a t^{2} \cdot \frac{b}{t} \cdot \frac{b}{t}}+b \cdot 3 \sqrt[3]{\frac{b}{t^{2}} \cdot a t \cdot a t}+a^{2}+b^{2} \\
=a^{2}+3 a^{\frac{4}{3}} b^{\frac{2}{3}}+3 a^{\frac{2}{3}} b^{\frac{4}{3}}+b^{2}=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{3} .
\end{array}
$$

The equality holds if and only if $a t^{2}=\frac{b}{t}, \frac{b}{t^{2}}=a t$, i.e., $t=\sqrt[3]{\frac{b}{a}}$.
Therefore, the minimum value of $(m+n)^{2}$ is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{3}$.
$$
\text { Hence }(m+n)_{\min }=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}} \text {. }
$$