Example 8 In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=a, B C=b, C C_{1}=c(a>b>c)$. Let the area of the section passing through $B D_{1}$ be $S$. Find the minimum value of $S$, and indicate the position of the section when $S$ is at its minimum (i.e., indicate the position of the intersection points of the section with the relevant edges).
(1) The sections $A B C_{1} D_{1}$, $B C D_{1} A_{1}$, and $D B B_{1} D_{1}$ are all rectangles, with their areas denoted as $S_{1}$, $S_{2}$, and $S_{3}$, respectively, then
$$
\begin{array}{l}
S_{1}=a \sqrt{b^{2}+c^{2}}, S_{2}=b \sqrt{c^{2}+a^{2}}, \\
S_{3}=c \sqrt{a^{2}+b^{2}} .
\end{array}
$$

Since $a>b>c$, it is easy to prove that $S_{1}>S_{2}>S_{3}$.
Therefore, we only need to consider the section $D B B_{1} D_{1}$.
(2) The section is a parallelogram, with three possible positions as shown in Figure 10(a), (b), (c) (sections $B E D_{1} F$, $B P D_{1} Q$, and $B R D_{1} S$), with their areas denoted as $S_{4}$, $S_{5}$, and $S_{6}$, respectively.
For the section $B E D_{1} F$, draw $E H \perp B D_{1}$ at point $H$ (as shown in Figure 10(a)), then $S_{4}=E H \cdot B D_{1}$.

Since $B D_{1}$ is a constant, when $E H$ takes its minimum value, $S_{4}$ has its minimum value $S_{4}^{\prime}$.

When $E H$ is the common perpendicular of the skew lines $B D_{1}$ and $A A_{1}$, it has its minimum value, which is the distance from point $A_{1}$ to the plane $D B B_{1} D_{1}$, i.e., the altitude of the right triangle $\triangle A_{1} B_{1} D_{1}$ on the hypotenuse $B_{1} D_{1}$, which is $\frac{a b}{\sqrt{a^{2}+b^{2}}}$. Therefore,
$$
S_{4}^{\prime}=\frac{a b}{\sqrt{a^{2}+b^{2}}} \cdot \sqrt{a^{2}+b^{2}+c^{2}} .
$$

Similarly, $S_{5}^{\prime}=\frac{a c}{\sqrt{a^{2}+c^{2}}} \cdot \sqrt{a^{2}+b^{2}+c^{2}}$,
$$
S_{6}^{\prime}=\frac{b c}{\sqrt{c^{2}+b^{2}}} \cdot \sqrt{a^{2}+b^{2}+c^{2}} \text {. }
$$

It is easy to prove that $S_{4}^{\prime}>S_{5}^{\prime}>S_{6}^{\prime}$ and $S_{6}^{\prime}<S_{3}$.
Therefore, the section $B R D_{1} S$ has the smallest area, with the minimum value being
$$
\frac{b c}{\sqrt{b^{2}+c^{2}}} \cdot \sqrt{a^{2}+b^{2}+c^{2}} \text {. }
$$

At this time, $B_{1} R=O G=\frac{a c^{2}}{b^{2}+c^{2}}$ (as shown in Figure 10(d)).