Example 11 As shown in Figure 2.10.10, through any point $P$ inside $\triangle ABC$, draw three lines parallel to its sides, forming three triangles with two of these lines and one side of the triangle. Let the areas of these three triangles be $S_{1}, S_{2}, S_{3}$. Try to find such a point that $S_{1}+S_{2}+S_{3}$ reaches its minimum value.
Let $A B=c, D P=A L=u, G P=B K=v$, then $L K=c-u-v$. Also,
$$
\begin{array}{c}
S_{\triangle D P E}=\frac{u^{2} \cdot \sin A \cdot \sin B}{2 \sin ^{2} C}=S_{1} \\
S_{\triangle F \mu G}=\frac{v^{2} \cdot \sin A \cdot \sin B}{2 \sin ^{2} C}=S_{2} \\
S_{\triangle P L X}=\frac{(c-u-v)^{2} \cdot \sin A \cdot \sin B}{2 \sin ^{2} C}=S_{3}
\end{array}
$$

Thus,
$$
\begin{aligned}
S_{1}+S_{2}+S_{3}= & \frac{\sin A \cdot \sin B}{2 \sin ^{2} C}\left[u^{2}+v^{2}+(c-u-v)^{2}\right] \geqslant \\
& \frac{3 \sin A \cdot \sin B}{2 \sin ^{2} C}\left[\frac{u+v+(c-u-v)}{3}\right]^{2} \\
& \text { (Quadratic mean and arithmetic mean inequality) }= \\
& \frac{c^{2} \sin A \cdot \sin B}{6 \sin ^{2} C}
\end{aligned}
$$

The equality holds if and only if $u=v=c-u-v=\frac{c}{3}$. 
Therefore, when $P$ is the centroid of $\triangle A B C$, $S_{1}+S_{2}+S_{3}$ achieves its minimum value.