Example 8 Given an equilateral $\triangle ABC$ with side length 4, points $D, E, F$ are on $BC, CA, AB$ respectively, and $|AE|=|BF|=|CD|=1$. Connecting $AD, BE, CF$ intersect each other to form $\triangle RQS$. Point $P$ moves within $\triangle PQR$ and on its boundary, and the distances from $P$ to the three sides of $\triangle ABC$ are $x, y, z$.
(1) Prove that when point $P$ is at the vertices of $\triangle RQS$, $xyz$ has a minimum value;
(2) Find the minimum value of the above $xyz$.


Solution: (1) Let $P$ be any point inside or on the sides of the equilateral $\triangle ABC$. Draw a line through $P$ parallel to $BC$ intersecting $AB$, $BE$, $AD$, and $AC$ at $U$, $V$, $M$, and $N$ respectively, as shown in Figure 20-2 (1). When point $P$ moves along $UN$, since the sum of the distances from $P$ to each side is a constant $x+y+z$ (the height of one side of the equilateral triangle), when the product $xyz$ has a fixed $x$, by $yz = PU \cdot PN \sin \frac{2\pi}{3}$, we get that $yz$ changes in the same way as $PU \cdot PN$.

Since $PU \cdot PN = \frac{1}{4} \left[(PU + PN)^2 - (PU - PN)^2\right]$, and $PU + PN$ is a constant, when $|PU - PN|$ gradually increases, $PU \cdot PN$ gradually decreases; when $P$ coincides with $G$ (the intersection of the height $AH$ on side $BC$ and $UN$), $PU \cdot PN$ reaches its maximum value. When $P$ moves away from $G$ to both sides, $PU \cdot PN$ gradually decreases. Thus, to make $xyz$ reach its minimum value, first adjust point $P$ to points $V$ and $M$, i.e., to the boundary of $\triangle RQS$.
Draw $RR' \parallel SQ' \parallel BC$, $QR' \parallel SS' \parallel AC$, and $QQ' \parallel RS' \parallel AB$, forming the hexagon $QQ'SS'RR'$, as shown in Figure 20-2 (2). In this way, we can repeat the above adjustments, moving point $P$ to points $P'$ and $P''$, and then continue to points $Q(Q')$ or $S(S')$. Due to symmetry, the value of $xyz$ at points $Q$, $Q'$, $S$, and $S'$ is the same. Similarly, the value of $xyz$ at points $R$ and $R'$ is the same as at point $Q$, so $xyz$ reaches its minimum value at points $Q$, $S$, and $R$.
(2) Since $\triangle ASE \sim \triangle ACD$, we have $AS:SE = 4:1$. Also, $\triangle AFR \sim \triangle ADB$, so $\frac{AR}{RF} = \frac{4}{3}$. Therefore, $AS:SR:RD = 4:8:1$. Given that the height of $\triangle ABC$ is $2\sqrt{3}$, we have $(xyz)_4 = \left(\frac{1}{13} \cdot 2\sqrt{3}\right)\left(\frac{9}{13} \cdot 2\sqrt{3}\right)\left(\frac{3}{13} \cdot 2\sqrt{3}\right) = \frac{648}{2197} \sqrt{3}$. Thus, the minimum value sought is $\frac{648}{2197} \sqrt{3}$.