Example 1. A spider is at point $A$ of the rectangular prism $A C^{\prime}$, and a fly is at point $C^{\prime}$. How can the spider reach the fly by the shortest route?
As shown in the figure, $AB$
$$
=a, BC=3, CC^{\prime}=
$$
$c$, and without loss of generality, let $a>b>c$.
Then, there are three possible shortest routes from $A \rightarrow C^{\prime}$:
(1) From the front
$A B B^{\prime} A^{\prime} \rightarrow$ to the top $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ (or bottom
to the back),
the shortest distance is
$$
\begin{array}{l}
\sqrt{a^{2}+(b+c)^{2}} \\
=\sqrt{a^{2}+b^{2}+c^{2}+2 b c} .
\end{array}
$$
(2) From the bottom $A B C D \rightarrow$ to the right $B C C^{\prime} B^{\prime}$ (or left to the top):
the shortest distance is
$$
\begin{array}{l}
\sqrt{(a+c)^{2}+b^{2}} \\
=\sqrt{a^{2}+b^{2}+c^{2}+2 a c} .
\end{array}
$$
(3) From the left $A A^{\prime} D^{\prime} D \rightarrow$ to the back $D D^{\prime} C^{\prime} C$ (or front to the right),
the shortest distance is
$$
\begin{aligned}
& \sqrt{(a+b)^{2}+c^{2}} \\
= & \sqrt{a^{2}+b^{2}+c^{2}+2 a b}
\end{aligned}
$$
Comparing (1), (2), and (3), (1) is the shortest.