[Example 1.3.7] A sphere with radius $r$ is inscribed in a regular tetrahedron. Find the minimum total surface area of this regular tetrahedron.
Regarding the problem of a sphere being tangent to a pyramid, it is generally necessary to select an appropriate section. Through this section, the problem of the sphere being tangent to the pyramid is reduced to the problem of a circle being tangent to a triangle, thus simplifying the problem.

As shown in the figure, a section is made through the height $P O_{1}$ of the regular triangular pyramid and a side edge $P C$, intersecting the edge $A B$ at its midpoint $D$. We focus our problem-solving excitement on the section $P C D$.

On this section, the great circle $O$ of the sphere is tangent to $P D$ and $C D$, and $O D$ bisects $\angle P D C$. Let $\angle P D C=\theta$, then
$$
\begin{array}{c}
D O_{1}=\frac{r}{\tan \frac{\theta}{2}}, \\
P D=\frac{D O_{1}}{\cos \theta}=\frac{r\left(1+\tan ^{2} \frac{\theta}{2}\right)}{\tan \frac{\theta}{2}\left(1-\tan ^{2} \frac{\theta}{2}\right)}, \\
A B=2 \sqrt{3} D O_{1}=\frac{2 \sqrt{3} r}{\tan \frac{\theta}{2}},
\end{array}
$$

Let the total surface area of the regular triangular pyramid be $S$, then
$$
\begin{array}{l}
S=\frac{\sqrt{3}}{4} A B^{2}+3 \cdot \frac{1}{2} A B \cdot P D \\
=\frac{\sqrt{3}}{4} \cdot \frac{12 r^{2}}{\tan ^{2} \frac{\theta}{2}}+\frac{3}{2} \cdot \frac{2 \sqrt{3} r^{2}\left(1+\tan ^{2} \frac{\theta}{2}\right)}{\tan ^{2} \frac{\theta}{2}\left(1-\tan ^{2} \frac{\theta}{2}\right)} \\
=\frac{6 \sqrt{3} r^{2}}{\tan ^{2} \frac{\theta}{2}\left(1-\tan ^{2} \frac{\theta}{2}\right)} \\
\geqslant \frac{6 \sqrt{3} r^{2}}{\left(\frac{\tan ^{2} \frac{\theta}{2}+1-\tan ^{2} \frac{\theta}{2}}{2}\right)^{2}} \\
=24 \sqrt{3} r^{2} .
\end{array}
$$

When and only when $\tan ^{2} \frac{\theta}{2}=1-\tan ^{2} \frac{\theta}{2}, \frac{\theta}{2}=\arctan \frac{\sqrt{2}}{2}$, i.e., $\theta=2 \operatorname{arc} \tan \frac{\sqrt{2}}{2}$, the total surface area of the regular triangular pyramid has a minimum value of $24 \sqrt{3} r^{2}$.