Example 9 Let $a \leqslant b<c$ be the side lengths of a right triangle $ABC$, find the largest constant $M$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{M}{a+b+c}$ always holds.
When $a=b=\frac{\sqrt{2}}{2} c$, i.e., $\triangle A B C$ is an isosceles right triangle, we have
$$M \leqslant 2+3 \sqrt{2}$$

Below we prove that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{2+3 \sqrt{2}}{a+b+c}$$,
which is equivalent to
$$a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \geqslant(2+3 \sqrt{2}) a b c$$

always holds.
$$\begin{aligned}
\text { The left side of the above equation } & =c\left(a^{2}+b^{2}\right)+\left(a^{2} b+b \cdot \frac{c^{2}}{2}\right)+\left(a b^{2}+\frac{c^{2}}{a \cdot 2}\right)+\frac{1}{2} c^{2}(a+b) \\
& \geqslant 2 a b c+\sqrt{2} a b c+\sqrt{2} a b c+\frac{1}{2} c \cdot \sqrt{a^{2}+b^{2}} \cdot(a+b) \\
& \geqslant 2 a b c+2 \sqrt{2} a b c+\frac{1}{2} c \cdot \sqrt{2 a b} \cdot 2 \sqrt{a b} \\
& =(2+3 \sqrt{2}) a b c .
\end{aligned}$$

Therefore,
$$M_{\max }=2+3 \sqrt{2}$$