Example 8 Given the three sides of $\triangle ABC$ are $BC=a, CA=b, AB=c$, and $\triangle DEF$ is any inscribed triangle in $\triangle ABC$, try to express the minimum value of the sum of the squares of the three sides of $\triangle DEF$ in terms of $a, b, c$.

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First, we prove the following conclusion: If $G$ is any point inside $\triangle ABC$, and the distances from $G$ to the three sides $BC$, $CA$, $AB$ are $x$, $y$, $z$ respectively, then when $x: y: z = a: b: c$, the minimum value of $x^2 + y^2 + z^2$ is $\frac{4 S_{\triangle ABC}^2}{a^2 + b^2 + c^2}$.

In fact, by the Cauchy-Schwarz inequality, $\left(a^2 + b^2 + c^2\right)\left(x^2 + y^2 + z^2\right) \geq (ax + by + cz)^2 = 4 S_{\triangle ABC}^2$. Equality holds if and only if $x: y: z = a: b: c$, thus proving the conclusion.

As shown in Figure 14-13, let $G$ be the centroid of $\triangle DEF$. Then, by the median length formula or centroid property 3(2), we have:
$$
\begin{array}{l}
G D^2 = \frac{1}{9}\left[2\left(D E^2 + D F^2\right) - E F^2\right], \\
G E^2 = \frac{1}{9}\left[2\left(D F^2 + E F^2\right) - D F^2\right], \\
G F^2 = \frac{1}{9}\left[2\left(E F^2 + D F^2\right) - D E^2\right].
\end{array}
$$

Adding these three equations, we get $D E^2 + E F^2 + F D^2 = 3\left(G D^2 + G E^2 + G F^2\right)$.
From point $G$, draw perpendiculars to the three sides $BC$, $AC$, $AB$ of $\triangle ABC$, with the feet of the perpendiculars being $D_0$, $E_0$, $F_0$ respectively. Then, $D E^2 + E F^2 + F D^2 = 3\left(G D_0^2 + G E_0^2 + G F_0^2\right) + 3\left(D D_0^2 + E E_0^2 + F F_0^2\right) \geq 3\left(G D_0^2 + G E_0^2 + G F_0^2\right) \geq \frac{12 S_{\triangle ABC}^2}{a^2 + b^2 + c^2}$.

Next, we prove that the equality can be achieved. Let $G$ be a point inside $\triangle ABC$, and the distances from $G$ to $BC$, $CA$, $AB$ are $x$, $y$, $z$ respectively, and satisfy $x: y: z = a: b: c$. Draw perpendiculars from $G$ to the three sides, with the feet of the perpendiculars being $D_0$, $E_0$, $F_0$. Since $D_0$, $C$, $E_0$, $G$ are concyclic, $\angle D_0 G E_0 + \angle C = 180^\circ$, thus $\frac{S_{\triangle G D_0 E_0}}{S_{\triangle ABC}} = \frac{\frac{1}{2} x y \cdot \sin \angle D_0 G E_0}{\frac{1}{2} a b \cdot \sin \angle C} = \frac{x y}{a b}$.
Similarly, $\frac{S_{\triangle G E_0 F_0}}{S_{\triangle ABC}} = \frac{y z}{b c}$, $\frac{S_{\triangle G F_0 D_0}}{S_{\triangle ABC}} = \frac{z x}{c a}$.
Since $x: y: z = a: b: c$, then $\frac{x y}{a b} = \frac{y z}{b c} = \frac{z x}{c a}$, hence $S_{\triangle G D_0 E_0} = S_{\triangle G E_0 F_0} = S_{\triangle G F_0 D_0}$. By centroid property 4(1), $G$ is the centroid of $\triangle D_0 E_0 F_0$. Therefore, for the inscribed $\triangle D_0 E_0 F_0$ of $\triangle ABC$, $D_0 E_0^2 + E_0 F_0^2 + F_0 D_0^2 = \frac{12 S_{\triangle ABC}^2}{a^2 + b^2 + c^2}$.
Thus, the minimum value sought is $\frac{12 S_{\triangle ABC}^2}{a^2 + b^2 + c^2}$.