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

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15. Since the ellipse $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1$ passes through the point $P(a, b)$, we have $\frac{a^{2}}{m^{2}}+\frac{b^{2}}{n^{2}}=1$. Therefore,
$$
\left(\frac{a}{m}\right)^{2}+\left(\frac{b}{m}\right)^{2}=1
$$

Since $a b \neq 0$, by the symmetry of the ellipse, we can assume $a, b \in \mathbf{R}^{+}$, and also $m, n \in \mathbf{R}^{+}$. Hence, we can set
$$
\left\{\begin{array}{l}
\frac{a}{m} = \cos \alpha, \\
\frac{b}{n} = \sin \alpha,
\end{array} \text{ i.e., } \left\{\begin{array}{l}
m=\frac{a}{\cos \alpha}, \\
n=\frac{b}{\sin \alpha},
\end{array} \text{ ( } \alpha \text{ is an acute angle) }
\right.
$$

Thus, $m+n=\frac{a}{\cos \alpha}+\frac{b}{\sin \alpha}$,
Let $t=\tan \alpha(t>0)$, then we get
$$
\begin{array}{l}
(m+n)^{2}=a^{2} t^{2}+\frac{b^{2}}{t^{2}}+2 a b t+\frac{2 a b}{t}+a^{2}+b^{2} \\
=a\left(a t^{2}+\frac{b}{t}+\frac{b}{t}\right)+b\left(\frac{b}{t^{2}}+a t+a t\right)+a^{2}+b^{2} \\
\geqslant a \cdot 3 \sqrt[3]{a t^{2} \cdot \frac{b}{t} \cdot \frac{b}{t}}+b \cdot 3 \sqrt[3]{\frac{b}{t^{2}} \cdot a t \cdot a t}+a^{2}+b^{2} \\
=a^{2}+3 a^{\frac{1}{3}} b^{\frac{2}{3}}+3 a^{\frac{2}{3}} b^{\frac{4}{3}}+b^{2} \\
=\left(a^{\frac{2}{4}}+b^{\frac{2}{4}}\right)^{3} \text {, } \\
\end{array}
$$

Equality holds if and only if $\left\{\begin{array}{l}a t^{2}=\frac{b}{t}, \\ \frac{b}{t^{2}}=a t\end{array}\right.$, i.e., $t=\sqrt[3]{\frac{b}{a}}$. Therefore, the minimum value of $(m+n)^{2}$ is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{3}$, and the minimum value of $(m+n)$ is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{1}{3}}$.