Proposition: Let the area of $\triangle ABC$ be $\triangle$, and the lengths of its three sides be $a$, $b$, and $c$. Then the minimum area of the inscribed equilateral triangle in $\triangle ABC$ is $\frac{\Delta^{2}}{\frac{\sqrt{3}}{6}\left(a^{2}+b^{2}+c^{2}\right)+2 \triangle}$.
Prove: As shown in Figure 1, the equilateral $\triangle PQR$ is inscribed in
$$
\begin{array}{l}
\triangle ABC, BC=a, CA \\
=b, AB=c.
\end{array}
$$

Let $\angle BRP=\theta$, then it is easy to find that $\angle PQC=\angle A + 60^\circ - \theta$. Let the side length of $\triangle PQR$ be $x$, then by the Law of Sines in $\triangle BRP$ and $\triangle PQC$, we get $BP = \frac{\sin \theta}{\sin B} x, PC = \frac{\sin (\angle A + 60^\circ - \theta)}{\sin C} x$.
Since $BP + PC = BC = a$, we have
$$
\begin{array}{l}
x = \frac{a}{\frac{\sin \theta}{\sin B} + \frac{\sin (\angle A + 60^\circ - \theta)}{\sin C}} \\
= \frac{a}{\frac{\sin (\angle A + 60^\circ)}{\sin C} \cdot \cos \theta + \left[\frac{1}{\sin B} - \frac{\cos (\angle A + 60^\circ)}{\sin C}\right] \cdot \sin \theta} \\
\geq \frac{a}{\sqrt{\left[\frac{\sin (\angle A + 60^\circ)}{\sin C}\right]^2 + \left[\frac{1}{\sin B} - \frac{\cos (\angle A + 60^\circ)}{\sin C}\right]^2}} \\
= \frac{a \sin B \cdot \sin C}{\sqrt{\sin^2 B + \sin^2 C + \sin B \cdot \sin C (\sqrt{3} \sin A - \cos A)}}. \\
\end{array}
$$

Substituting $\sin A = \frac{2 \Delta}{bc}, \sin B = \frac{2 \Delta}{ca}, \sin C = \frac{2 \Delta}{ab}, \cos A = \frac{b^2 + c^2 - a^2}{2bc}$, we get

And it is easy to see that when $\tan \theta = \frac{\sin C - \sin B \cdot \cos (\angle A + 60^\circ)}{\sin B \cdot \sin (\angle A + 60^\circ)}$, the equality can be achieved, hence the minimum area of the equilateral $\triangle PQR$ is $\frac{\sqrt{3}}{4} \cdot \frac{8 \Delta^2}{a^2 + b^2 + c^2 + 4 \sqrt{3} \Delta} = \frac{\Delta^2}{\frac{\sqrt{3}}{6}(a^2 + b^2 + c^2) + 2 \Delta}$. Specifically, when $\triangle ABC$ is an equilateral triangle, the minimum area of the equilateral $\triangle PQR$ is $\frac{1}{4} \Delta$; when $\triangle ABC$ is a right triangle (with $\angle C$ being the right angle), the minimum area of the equilateral $\triangle PQR$ is
$$
\frac{1}{\frac{4 \sqrt{3}}{3}\left(\frac{1}{a^2} + \frac{1}{b^2}\right) + \frac{-4}{ab}}.
$$