Given an equilateral triangle $ABC$ with side length 4. Points $D$, $E$, and $F$ are on $BC$, $CA$, and $AB$ respectively, and $|AE|=|BF|=|CD|=1$. Connect $AD$, $BE$, and $CF$, intersecting to form $\triangle RQS$. Point $P$ moves within $\triangle RQS$ and on its sides. The distances from $P$ to the three sides of $\triangle ABC$ are denoted as $x$, $y$, and $z$,
(1) Prove that when point $P$ is at the vertices of $\triangle RQS$, the product $xyz$ has a minimum value;
(2) Find the minimum value of the product $xyz$.
Given the assumptions $A B=B C=C A=4, A E=B F=C D=1$.
$\therefore \quad \triangle A B E \cong \triangle B C F \cong \triangle C A D \quad(\mathrm{~S}, \mathrm{~A}, \mathrm{~S})$.
The intersection points of $A D$, $B E$, and $C F$ are denoted as $R$, $Q$, and $S$ as shown in the figure.
$\therefore \quad \triangle A E R \cong \triangle B F Q \cong \triangle C D S \quad(\mathrm{~A}, \mathrm{~S}, \mathrm{~A})$.
Thus, $\triangle R Q S$ is an equilateral triangle (all internal angles are equal).
Also, $A R=B Q=C S, D S=E R=F Q$.
For any point $P$ inside $\triangle A B C$, let $P_{1}, P_{2}, P_{3}$ denote the distances from $P$ to $B C$, $C A$, and $A B$ respectively. Thus, $R_{1}$, $R_{2}$, $R_{3}$ denote the distances from $R$ to $B C$, $C A$, and $A B$ respectively, $Q_{1}$, $Q_{2}$, $Q_{3}$ denote the distances from $Q$ to $B C$, $C A$, and $A B$ respectively, and $S_{1}$, $S_{2}$, $S_{3}$ denote the distances from $S$ to $B C$, $C A$, and $A B$ respectively, and so on.
It is easy to prove: $S_{1}=R_{2}=Q_{3}, S_{2}=R_{3}=Q_{1}, S_{3}=R_{1}=Q_{2}$, $\therefore \quad S_{1} S_{2} S_{3}=R_{1} R_{2} R_{3}=Q_{1} Q_{2} Q_{3}$

Furthermore, for any point $P$ inside the equilateral $\triangle A B C$, the sum of the distances from $P$ to the three sides is equal to the height of the equilateral $\triangle A B C$ (a constant). If two positive numbers $x, y$ have a sum of $a$, then their product
$$
\begin{aligned}
x y & =\frac{(x+y)^{2}}{4}-\frac{(x-y)^{2}}{4} \\
& =\frac{a^{2}}{4}-\frac{1}{4}(x-y)^{2} .
\end{aligned}
$$

Clearly, when $x=y=\frac{a}{2}$, $x y$ is maximized, and when $|x-y|$ is larger, $x y$ is smaller. (2)

First, fix the distance from $P$ to one side of the equilateral $\triangle A B C$, and examine the changes in $P_{1} P_{2} P_{3}$ as $P$ moves along the line segment perpendicular to this side and bisected by the altitude, as shown in the figure. The altitude $A H \perp M N$ at $P_{0}$, and $P_{0} M=P_{0} N$. By symmetry,
$$
M_{1} M_{2} M_{3}=N_{1} N_{2} N_{3} \text {. }
$$

According to (2), when $P$ coincides with $P_{0}$, $P_{1} P_{2} P_{3}$ is maximized, and when $P P_{0}$ is larger, $P_{1} P_{2} P_{3}$ is smaller.
$$
\therefore \quad P_{1} P_{2} P_{3} \geqslant M_{1} M_{2} M_{3}=N_{1} N_{2} N_{3},
$$

i.e., when $P$ moves along $M N$, the product $P_{1} P_{2} P_{3}$ reaches its minimum value at the endpoints $M$ and $N$.

Construct the symmetric points of $E$, $D$, $F$, $Q$, $R$, and $S$ with respect to the altitude $A H$: as shown in the figure, connect the relevant line segments to form the hexagon $\Omega: R^{\prime} R S^{\prime} S Q^{\prime} Q$. The three altitudes of $\triangle A B C$ are exactly the symmetry axes of $\Omega$, and each pair of opposite sides of the hexagon is parallel to one side of the equilateral $\triangle A B C$ and is bisected perpendicularly by the altitude of this side. By (1) and symmetry, it is easy to see:
$$
\begin{aligned}
R_{1} R_{2} R_{3} & =S^{\prime}{ }_{1} S^{\prime}{ }_{2} S^{\prime}{ }_{3}=S_{1} S_{2} S_{3}=Q^{\prime}{ }_{1} Q^{\prime}{ }_{2} Q^{\prime}{ }_{3} \\
& =Q_{1} Q_{2} Q_{3}=R^{\prime}{ }_{1} R^{\prime}{ }_{2} R^{\prime}{ }_{3} .
\end{aligned}
$$

Let $P$ be any point inside the hexagon $\Omega$, and draw a line $l \parallel B C$ intersecting the sides $R^{\prime} Q$ and $R S^{\prime}$ of $\Omega$ at $M$ and $N$ respectively (the case of intersecting $Q Q^{\prime}$ and $S^{\prime} S$ is similar). Since $M N$ is bisected perpendicularly by the altitude of $B C$, according to (3), we have $P_{1} P_{2} P_{3} \geqslant M_{1} M_{2} M_{3}=N_{1} N_{2} N_{3}$. Since $N$ is a point on the line segment $R S^{\prime}$, and $R S^{\prime}$ is bisected perpendicularly by the altitude of $A C$, according to (3), we have $N_{1} N_{2} N_{3} \geqslant R_{1} R_{2} R_{3}$.
$\therefore \quad P_{1} P_{2} P_{3} \geqslant R_{1} R_{2} R_{3}$. Noting (4), we know that when $P$ varies inside the hexagon $\Omega$ (including the sides), the minimum value is taken at the six vertices, but $\triangle R Q S$ is contained within the hexagon $\Omega$, and its vertices $R$, $Q$, and $S$ are exactly three vertices of the hexagon $\Omega$, so when $P$ varies inside $\triangle R Q S$, $P_{1} P_{2} P_{3}$ takes its minimum value at the three vertices $R$, $Q$, and $S$. Now, we calculate the minimum value, for example, by calculating $S_{1} S_{2} S_{3}$.
By
$\therefore$
Since
Since
$\qquad$Therefore,
Therefore
Since the height of $\triangle A B C$ is
$$
\begin{array}{l}
h=\sqrt{12} \text {. } \\
\therefore \quad S_{1}=\frac{1}{13} h, \quad S_{3}=R_{1}=\frac{9}{13} h, \quad S_{2}=\frac{3}{13} h . \\
\therefore \quad S_{1} S_{2} S_{3}=\frac{1}{13} \cdot \frac{3}{13} \cdot \frac{9}{13}(\sqrt{12})^{3}=\frac{648}{2197} \sqrt{3} . \\
\end{array}
$$