(6) In an isosceles right triangle $\triangle ABC$, $CA = CB = 1$, and point $P$ is any point on the boundary of $\triangle ABC$. Find the maximum value of $PA \cdot PB \cdot PC$. (Li Weiguo)
Hui Fa Yi First prove that the maximum value can only be obtained when $P \in AB$.
(1) When $P \in AC$, we have
$$
PA \cdot PC \leqslant \frac{1}{4}, PB \leqslant \sqrt{2},
$$

Thus,
$$
PA \cdot PB \cdot PC \leqslant \frac{\sqrt{2}}{4},
$$

where the equality does not hold (since the two equalities cannot hold simultaneously), i.e.,
$$
PA \cdot PB \cdot PC < \frac{\sqrt{2}}{4}.
$$
(2) When $P \in AB$, let $AP = x \in [0, \sqrt{2}]$, then
$$
\begin{aligned}
f(x) &= PA^2 \cdot PB^2 \cdot PC^2 \\
&= x^2(\sqrt{2}-x)^2\left(1 + x^2 - \sqrt{2} x\right).
\end{aligned}
$$

Let $t = x(\sqrt{2} - x)$, then $t \in \left[0, \frac{1}{2}\right]$, and $f(x) = g(t) = t^2(1 - t)$. Note that $g'(t) = 2t - 3t^2 = t(2 - 3t)$, so $g(t)$ is increasing on $\left[0, \frac{2}{3}\right]$. Therefore, $f(x) \leq g\left(\frac{1}{2}\right) = \frac{1}{8}$, hence $PA \cdot PB \cdot PC \leq \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$. The equality holds if and only if $t = \frac{1}{2}$, $x = \frac{\sqrt{2}}{2}$, i.e., when $P$ is the midpoint of $AB$.

Hui Fa Er (1) When $P$ is on the line segment $AB$, draw $CD \perp AB$ at $D$.
Let $PD = a$, then
$$
PA \cdot PB = \frac{1}{2} - a^2, PC = \sqrt{\frac{1}{2} + a^2},
$$

Thus,
$$
\begin{aligned}
PA \cdot PB \cdot PC &= \left(\frac{1}{2} - a^2\right) \cdot \sqrt{\frac{1}{2} + a^2} \\
&\leqslant \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} - a^2} \cdot \sqrt{\frac{1}{2} + a^2} \\
&\leqslant \frac{\sqrt{2}}{2} \cdot \sqrt{\frac{1}{4}} \\
&= \frac{\sqrt{2}}{4}
\end{aligned}
$$

The equality holds when $a = 0$.
(2) When $P$ is on the hypotenuse, assume $P$ is on $BC$, let $PC = b$, then
$$
\begin{aligned}
PA \cdot PB \cdot PC &= b(1 - b) \sqrt{1 + b^2} \\
&\leqslant \left(\frac{b + 1 - b}{2}\right)^2 \sqrt{1 + b^2} \\
&= \frac{1}{4} \sqrt{1 + b^2} \leqslant \frac{1}{4} \sqrt{1 + 1^2} \\
&= \frac{\sqrt{2}}{4}
\end{aligned}
$$

The equality clearly cannot hold.
Therefore, the maximum value of $PA \cdot PB \cdot PC$ is $\frac{\sqrt{2}}{4}$. The equality holds when $P$ is the midpoint of $AB$.