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 $z$ respectively.
(1) Prove that when point $P$ is at the vertices of $\triangle RQS$, the product $xyz$ has a minimum value; $\square$
(2) Find the minimum value of the product $xyz$.
Given the assumptions $A B=B C=C A=4, A E=B F=C D=1$,
$\therefore \triangle A B E \cong \triangle B C F \cong \triangle C A D(s, a, s)$.
The intersection points of $A D$, $B E$, and $C F$ are denoted as $R$, $Q$, and $S$ as shown in the figure.
$\therefore \triangle A E R \cong \triangle B F Q \cong \triangle C D S(a, s, a)$.
Thus, $\triangle R Q S$ is an equilateral triangle
(all internal angles are equal).
Also, $A R=B Q=C S$,
$$
D S=E R=F Q \text {. }
$$

For any point $P$ inside $\triangle A B C$,
let $P_{1}, P_{2}, P_{3}$
represent the distances from point $P$ to $B C$, $C A$, and $A B$, respectively. Thus, $R_{1}, R_{2}, R_{3}$
represent the distances from point $R$ to $B C$, $C A$, and $A B$, $Q_{1}, Q_{2}, Q_{3}$ represent the distances from point $Q$ to $B C$, $C A$, and $A B$, and $S_{1}, S_{2}, S_{3}$ represent the distances from point $S$ to $B C$, $C A$, and $A B$, respectively. The rest are similarly denoted.
$$
\text { It is easy to prove: } \quad \begin{array}{l}
S_{1}=R_{2}=Q_{3}, \\
\\
S_{2}=R_{3}=Q_{1}, \\
S_{3}=R_{1}=Q_{2},
\end{array}
$$

It is easy to prove:
$$
\therefore \quad S_{1} S_{2} S_{2}=R_{1} R_{2} R_{3}=Q_{1} Q_{2} Q_{3}
$$

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 $x, y$ sum to $a$, then their product
$$
x y=\frac{(x+y)^{2}}{4}-\frac{(x-y)^{2}}{4}=\frac{a^{2}}{4}-\frac{1}{4}(x-y)^{2} .
$$

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

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,
$$
M_{1} M_{2} M_{3}=N_{1} N_{2} N_{3} .
$$

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.
$$
\therefore \quad P_{1} P_{2} P_{3} \geqslant M_{1} M_{2} M_{3}=N_{1} N_{2} N_{3} \text {, }
$$

That is, as $P$ 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 axes of symmetry 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 that side. By (1) and symmetry, it is easy to know:
$$
\begin{aligned}
R_{1} R_{2} R_{3} & =S^{\prime}{ }_{1} S^{\prime}{ }_{2} S^{\prime}{ }_{3}=S_{1} S_{2} S_{3}=Q^{\prime}{ }_{1} Q^{\prime}{ }_{2} Q^{\prime}{ }_{3} \\
& =Q_{1} Q_{2} Q_{3}=R^{\prime}{ }_{1} R_{2}^{\prime} R^{\prime}{ }_{3} .
\end{aligned}
$$

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 $N$ (the case of intersecting $Q Q^{\prime}$ and $S^{\prime} S$ is similarly proven). 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 $N_{1} N_{2} N_{3} \geqslant R_{1} R_{2} R_{3}$.
$$
\therefore \quad P_{1} P_{2} P_{3} \geqslant R_{1} R_{2} R_{3} \text {. }
$$

Noting (4), we know that as $P$ varies within the hexagon $\Omega$ (including the sides), the minimum value is attained at the six vertices. However, $\triangle R Q S$ is contained within the hexagon $\Omega$, and its vertices $R$, $Q$, and $S$ are exactly three vertices of the hexagon $\Omega$. Therefore, as $P$ varies within $\triangle R Q S$,
$P_{1} P_{2} P_{3}$ attains its minimum value at the three vertices $R$, $Q$, and $S$. We now calculate the minimum value, for example, by calculating $S_{1} S_{2} S_{3}$.
\begin{tabular}{ll} 
By & $\triangle A R E \backsim \triangle A D C$, \\
$\therefore$ & $A R: R E=4: 1$. \\
Since & $\triangle A F S \backsim \triangle A B D$, \\
$\therefore$ & $A S: S F=4: 3$,
\end{tabular}

Therefore, $\quad A R: R S: S D=4: 8: 1$.
Since the height of $\triangle A B C$ is
$$
\begin{array}{cc} 
& h=\sqrt{12} . \\
\therefore & S_{1}=\frac{1}{13} h, S_{3}=R_{1}=\frac{9}{13} h, S_{2}=\frac{3}{13} h . \\
\therefore & S_{1} S_{2} S_{3}=\frac{1}{13} \cdot \frac{3}{13} \cdot \frac{9}{13}(\sqrt{12})^{3}=\frac{648}{2197} \sqrt{3} .
\end{array}
$$