$9 \cdot 39$ Let $a \leqslant b<c$ be the side lengths of a right triangle, find the maximum constant $M$, such that
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{M}{a+b+c} .
$$
[Solution] Let $I=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$, then
$$
I=3+\frac{b}{a}+\frac{a}{b}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b} .
$$

Since
$$
\begin{array}{l}
\frac{b}{a}+\frac{a}{b} \geqslant 2, \\
\frac{a}{c}+\frac{c}{a}=\frac{2 a}{c}+\frac{c}{a}-\frac{a}{c} \geqslant 2 \sqrt{2}-\frac{a}{c}, \\
\frac{b}{c}+\frac{c}{b}=\frac{2 b}{c}+\frac{c}{b}-\frac{b}{c} \geqslant 2 \sqrt{2}-\frac{b}{c} .
\end{array}
$$

Therefore, $I \geqslant 5+4 \sqrt{2}-\frac{a+b}{c}$.
Also, $\quad \frac{a+b}{c} \leqslant \frac{\sqrt{2\left(a^{2}+b^{2}\right)}}{c}=\sqrt{2}$,
thus $\quad I \geqslant 5+3 \sqrt{2}$.
When $a=b=1, c=\sqrt{2}$, $I=(2+\sqrt{2})\left(2+\frac{\sqrt{2}}{2}\right)=5+3 \sqrt{2}$,
hence $M=5+3 \sqrt{2}$.