32. Let $a \leqslant b<c$ be the side lengths of a right triangle, find the largest constant $M$ such that $\frac{1}{a}+\frac{1}{b}+$ $\frac{1}{c} \geqslant \frac{M}{a+b+c}$. (1991 China National Training Team Test)
$$\begin{array}{l}
32 \text { Let } I=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+1+1=3+\frac{b}{a}+\frac{a}{b}+\frac{c}{a}+\frac{a}{c}+\frac{b}{c}+\frac{c}{b}\right. \\
\text { Since } \frac{b}{a}+\frac{a}{b} \geqslant 2, \frac{c}{a}+\frac{a}{c}=\frac{c}{a}+\frac{2 a}{c}-\frac{a}{c} \geqslant 2 \sqrt{2}-\frac{a}{c}, \frac{b}{c}+\frac{c}{b}=\frac{c}{b}+\frac{2 b}{c}-\text {. } \\
\frac{b}{c} \geqslant 2 \sqrt{2}-\frac{b}{c} \text {, therefore, } I \geqslant 5+4 \sqrt{2}-\frac{a+b}{c} \text {, and } \frac{a+b}{c} \leqslant \frac{\sqrt{2\left(a^{2}+b^{2}\right)}}{c}=\sqrt{2} \text {, thus } \\
i \geqslant 5+3 \sqrt{2} \\
\text { Second, when } a=b=1, c=\sqrt{2} \text {, } L=(2+\sqrt{2})\left(2+\frac{\sqrt{2}}{2}\right)=5+3 \sqrt{2} \text {, so the maximum } I= \\
5+3 \sqrt{2}
\end{array}$$