In an acute triangle $ABC$, $BC=a$, $AC=b$, $AB=c$. There are moving points $D$, $E$, $F$ on sides $BC$, $CA$, $AB$ respectively. Determine the area of $\triangle DEF$ when $DE^2 + EF^2 + FD^2$ is minimized.
For any inscribed $\triangle D E F$, temporarily fix $E F$, and let $D$ move on $B C$. Let the midpoint of $E F$ be $M$. Then, by the median length formula,
$$
D E^{2}+D F^{2}=\frac{E F^{2}}{2}+2 \cdot D M^{2},
$$
Therefore, when $E F$ is fixed, to minimize $D E^{2}+E F^{2}+F D^{2}$, $D M$ must be minimized. Since $M$ is a fixed point on $E F$, $D M$ is minimized when $D M \perp B C$. Similarly, for $E$ and $F$, it can be concluded that when $D E^{2}+E F^{2}+F D^{2}$ is minimized, the three medians of $\triangle D E F$ must be perpendicular to the corresponding sides of $\triangle A B C$. Let the centroid of $\triangle D E F$ be $G$, and its area be $S_{0}$, and the area of $\triangle A B C$ be $S$. Then,
$$
S_{\triangle G D E}=S_{\triangle G E F}=S_{\triangle G F D}=\frac{S_{0}}{3}.
$$

Since GDCE, GEAF, and GFBD are cyclic, then
$$
\begin{array}{l}
\angle D G E=\pi-C, \angle E G F=\pi-A, \\
\angle F G D=\pi-B,
\end{array}
$$

Thus, by (1),
$$
G D \cdot G E \cdot \sin C=G E \cdot G F \cdot \sin A=G F \cdot G D \cdot \sin B,
$$

Dividing each term by $2 S$, we get
$$
\frac{G D \cdot G E}{a \cdot b}=\frac{G E \cdot G F}{b \cdot c}=\frac{G D \cdot G F}{a \cdot c},
$$

Therefore,
$$
\frac{G D}{a}=\frac{G E}{b}=\frac{G F}{c}=\lambda,
$$

Also, by
$$
G D \cdot a+G E \cdot b+G F \cdot c=2 S,
$$

we have $\lambda\left(a^{2}+b^{2}+c^{2}\right)=2 S$, so $\lambda=\frac{2 S}{a^{2}+b^{2}+c^{2}}$. Therefore,
$$
\begin{aligned}
S_{0} & =3 \cdot \frac{1}{2} \cdot G D \cdot G E \sin C=3 \lambda^{2} \cdot \frac{1}{2} a b \sin C \\
& =3 \lambda^{2} S=\frac{12 S^{3}}{\left(a^{2}+b^{2}+c^{2}\right)^{2}}
\end{aligned}
$$
(Where $\left.S=\sqrt{p(p-a)(p-b)(p-c)}, p=\frac{a+b+c}{2}\right)$).