Given an equilateral triangle $ABC$ with side length 4. Points $D, E, F$ are on $BC, CA, AB$ respectively, and $|AE| = |BF| = |CD| = 1$. Connect $AD, BE, CF$, intersecting to form $\triangle RQS$. Point $P$ moves within $\triangle RQS$ and on its sides. The distances from point $P$ to the three sides of $\triangle ABC$ are denoted as $x, y, z$.

(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 $xyz$.
$\frac{648}{2197} \sqrt{3}$
IV. [Analysis and Solution]
Proof: Given $A B=B C=C A=4, A E=B F=C D=1$.
$\therefore \triangle A B E \cong \triangle B C F \cong \triangle C A D \quad(\mathrm{SAS})$.
The intersection points of $A D 、 E F 、 C F$ are $R 、 Q 、 S$, as shown in the figure.
$\therefore \triangle A E R \cong \triangle B F Q \cong \triangle C D S \quad$ (ASA).
Thus, $\triangle R Q S$ is an equilateral triangle (all three 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 $P_{1}, P_{2}, P_{3}$ represent the distances from point $P$ to $B C 、 C A 、 A B$, respectively. Thus, $R_{1}, R_{2}, R_{3}$ represent the distances from point $R$ to $B C 、 C A 、 A B$, and $Q_{1}, Q_{2}, Q_{3}$ represent the distances from point $Q$ to $B C 、 C A 、 A B$, and similarly for $S_{1}, S_{2}, S_{3}$.
It is easy to prove: $S_{1}=R_{2}=Q_{3}, S_{2}=R_{3}=Q_{1}, S_{3}=R_{1}=Q_{2}$,
$\therefore S_{1} S_{2} S_{3}=R_{1} R_{2} R_{3}=Q_{1} Q_{2} Q_{3}$
Furthermore, the sum of the distances from any point $P$ inside the equilateral $\triangle A B C$ 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 where the altitude $A H \perp M N$ at $P_{0}$, and $P_{0} M=P_{0} N$. By symmetry,
$\therefore M_{1} M_{2} M_{3}=N_{1} N_{2} N_{3}$, 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. $\therefore P_{1} P_{2} P_{3} \geqslant M_{1} M_{2} M_{3}=N_{1} N_{2} N_{3}$
That is, as $P$ moves along $M N$, the product $P_{1} P_{2} P_{3}$ reaches its minimum value at the endpoints $M 、 N$.
Construct the symmetric points of $E 、 D 、 F 、 Q 、 R 、 S$ about the altitude $A H$, as shown in the figure, and 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 see:
$R_{1} R_{2} R_{3}=S^{\prime}{ }_{1} S^{\prime}{ }_{2} S^{\prime}{ }_{3}=S_{1} S_{2} S_{3}=Q^{\prime}{ }_{1} Q^{\prime}{ }_{2} Q^{\prime}{ }_{3}=Q_{1} Q_{2} Q_{3}=R_{1}^{\prime} R^{\prime}{ }_{2} R^{\prime}{ }_{3}$.
Let $P$ be any point inside the hexagon $\Omega$, draw a line $l / / B C$ intersecting the sides $R^{\prime} Q 、 R S^{\prime}$ of $\Omega$ at $M 、 N$ (intersecting $Q Q^{\prime} 、 S^{\prime}$
(2)
$S$ 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 the midpoint of the segment $R S^{\prime}$ bisected perpendicularly by the altitude of $A C$, according to (3) we have $N_{1} N_{2} N_{3} \geqslant R_{1} R_{2} R_{3}$.
$\therefore P_{1} P_{2} P_{3} \geqslant R_{1} R_{2} R_{3}$. Noting (4), we see that as $P$ varies within the hexagon $\Omega$ (including the sides), it reaches its minimum value at the six vertices, but $\triangle R Q S$ is contained within the hexagon $\Omega$, and its vertices, so as $P$ varies within $\triangle R Q S$, $P_{1} P_{2} P_{3}$ reaches its minimum value at the three vertices $R 、 Q 、 S$. Below we calculate the minimum value, for example, by calculating $S_{1} S_{2} S_{3}$.
By $\triangle A R E \backsim A D C, \therefore A R: R E=4: 1$.
Since $\triangle A F S \sim \triangle A B D, \therefore A S: S F=4: 3$,
Therefore, $A R: R S: S D=4: 8: 1$.
Since the height $h$ of $\triangle A B C$ is $\sqrt{12} . \therefore S_{1}=\frac{1}{13} h, S_{3}=R_{1}=\frac{9}{13} h, S_{2}=\frac{3}{13} h$.
$$
\therefore S_{1} S_{2} S_{3}=\frac{1}{13} \cdot \frac{9}{13} \cdot \frac{3}{13}(\sqrt{12})^{3}=\frac{648}{2197} \sqrt{3} \text {. }
$$

[Extension] There are many methods to find the extremum of a multivariable function, and the adjustment method is one of them. This problem is a good example of using the adjustment method, which is worth repeated contemplation. In the adjustment process, "change" and "motion" can be seen, so the adjustment method is more vivid than other methods. The adjustment method is a common method for proving trigonometric inequalities, simple and easy to follow. For example:
"Prove: In triangle $A B C$, $\sin A+\sin B+\sin C \leqslant \frac{3 \sqrt{3}}{2}$." (Use the adjustment method. Assume $A \geqslant B \geqslant C$, keep $B$ unchanged, and adjust $A, C$ to $60^{\circ}$ and $A+C-60^{\circ}$.)