9. Given that the three sides of $\triangle A B C$ are $3, 4, 5$ respectively, and point $P$ is a moving point inside $\triangle A B C$ (excluding the boundary), then the maximum value of the product of the distances from point $P$ to the three sides is $\qquad$ .
9 $\frac{16}{15}$ Hint: Let $a=3, b=4, c=5$, then $\triangle A B C$ is a right triangle, with its area $S_{\triangle A B C}=6$. Let the distances from point $P$ to the three sides $a, b, c$ be $h_{a}, h_{b}, h_{c}$, respectively, then
$$
a h_{a}+b h_{b}+c h_{c}=2 S_{\triangle A B C}=12,
$$

Therefore,
$$
\begin{aligned}
h_{a} h_{b} h_{c} & =\frac{a h_{a} \cdot b h_{b} \cdot c h_{c}}{a b c} \\
& \leqslant \frac{1}{a b c}\left(\frac{a h_{a}+b h_{b}+c h_{c}}{3}\right)^{3} \\
& =\frac{4^{3}}{60}=\frac{16}{15},
\end{aligned}
$$

Equality holds if and only if
$$
\left\{\begin{array}{l}
a h_{a}=b h_{b}=c h_{c} \\
a h_{a}+b h_{b}+c h_{c}=12,
\end{array}\right.
$$

which means
$$
S_{\triangle P A B}=S_{\triangle P B C}=S_{\triangle P C A},
$$

i.e., when $P$ is the centroid of $\triangle A B C$.