Example 11 Given that $x, y, z$ are real numbers, satisfying
$$
\begin{array}{c}
x=\sqrt{y^{2}-\frac{1}{16}}+\sqrt{z^{2}-\frac{1}{16}}, \\
y=\sqrt{z^{2}-\frac{1}{25}}+\sqrt{x^{2}-\frac{1}{25}}, \\
z=\sqrt{x^{2}-\frac{1}{36}}+\sqrt{y^{2}-\frac{1}{36}},
\end{array}
$$

and $x+y+z=\frac{m}{\sqrt{n}}\left(m, n \in \mathbf{Z}_{+}, \sqrt{n}\right.$ is the simplest radical form). Find $m+n$.
Solve as shown in Figure 11, construct an acute triangle $\triangle ABC$, such that
$$
\begin{array}{l}
BC=x, \\
CA=y, \\
AB=z .
\end{array}
$$

Draw perpendiculars from $A$, $B$, and $C$ to $BC$, $CA$, and $AB$ respectively, with the feet of the perpendiculars being $D$, $E$, and $F$. Let $AD=u$, $BE=v$, and $CF=w$. Then
$$
\begin{array}{l}
x=\sqrt{y^{2}-u^{2}}+\sqrt{z^{2}-u^{2}}, \\
y=\sqrt{z^{2}-v^{2}}+\sqrt{x^{2}-v^{2}}, \\
z=\sqrt{x^{2}-w^{2}}+\sqrt{y^{2}-w^{2}} .
\end{array}
$$

Consider the function
$$
f(u)=\sqrt{y^{2}-u^{2}}+\sqrt{z^{2}-u^{2}}.
$$

It is easy to see that $f(u)$ is strictly decreasing on $(0,+\infty)$. Hence $u=\frac{1}{4}$.
Similarly, $v=\frac{1}{5}$, $w=\frac{1}{6}$.
Let the area of $\triangle ABC$ be $S$. Then
$$
S=\frac{1}{2} AB \cdot CF=\frac{1}{2} BC \cdot AD=\frac{1}{2} CA \cdot BE,
$$

which means
$$
\begin{array}{l}
2 S=z w=x u=y v \\
\Rightarrow x=8 S, y=10 S, z=12 S \\
\Rightarrow x+y+z=30 S .
\end{array}
$$

By Heron's formula,
$$
\begin{array}{l}
S^{2}=15 S(15 S-8 S)(15 S-10 S)(15 S-12 S) \\
\Rightarrow S=\frac{1}{15 \sqrt{7}} \Rightarrow x+y+z=\frac{2}{\sqrt{7}} \\
\Rightarrow m+n=9 .
\end{array}
$$