6. In an isosceles right triangle \( \triangle ABC \), \( CA = CB = 1 \), and \( P \) is any point on the boundary of \( \triangle ABC \). Find the maximum value of \( PA \cdot PB \cdot PC \).
(Li Weiguo)
6. (1) As shown in Figure 5, when $P \in A C$, we have
$$
P A \cdot P C \leqslant \frac{1}{4}, \quad P B \leqslant \sqrt{2}.
$$

Therefore, $P A \cdot P B \cdot P C \leqslant \frac{\sqrt{2}}{4}$,
where the equality does not hold (since the two equalities cannot hold simultaneously), i.e., $P A \cdot P B \cdot P C < \frac{\sqrt{2}}{4}$.
(2) As shown in Figure 6, when $P \in A B$, let $A P = x \in [0, \sqrt{2}]$, then
$$
\begin{array}{l}
f(x) = P A^{2} \cdot P B^{2} \cdot P C^{2} \\
= x^{2}(\sqrt{2} - x)^{2}\left(1 + x^{2} - \sqrt{2} x\right).
\end{array}
$$

Let $t = x(\sqrt{2} - x)$, then
$$
t \in \left[0, \frac{1}{2}\right], \quad f(x) = g(t) = t^{2}(1 - t).
$$

Notice that $g'(t) = 2t - 3t^{2} = t(2 - 3t)$, so $g(t)$ is increasing on $\left[0, \frac{2}{3}\right]$. Then
$$
f(x) \leqslant g\left(\frac{1}{2}\right) = \frac{1}{8}.
$$

Therefore, $P A \cdot P B \cdot P C \leqslant \frac{1}{2 \sqrt{2}} = \frac{\sqrt{2}}{4}$.
The equality holds if and only if $t = \frac{1}{2}, x = \frac{\sqrt{2}}{2}$, i.e., when $P$ is the midpoint of $A B$.