Example 2 The sum of the 6 edge lengths of the tetrahedron $P-ABC$ is $l$, and $\angle APB=\angle BPC=\angle CPA=90^{\circ}$. Then the maximum volume of the tetrahedron is $\qquad$
(5th United States of America Mathematical Olympiad)
Analysis: Let $P A=a, P B=b, P C=c$, it is easy to see that
$$
l=a+b+c+\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}+
$$
$\sqrt{c^{2}+a^{2}}, V=\frac{1}{6} a b c$.

We can see that both of the above formulas are cyclic symmetric expressions about $a, b, c$. It is predicted that the volume is maximized when $a=b=c$.
In fact,
$$
\begin{aligned}
l= & a+b+c+\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}+ \\
& \sqrt{c^{2}+a^{2}} \\
\geqslant & 3 \sqrt[3]{a b c}+ \\
& 3 \sqrt[3]{\sqrt{a^{2}+b^{2}} \cdot \sqrt{b^{2}+c^{2}} \cdot \sqrt{c^{2}+a^{2}}} \\
\geqslant & 3 \sqrt[3]{a b c}+3 \sqrt{2} \cdot \sqrt[3]{a b c},
\end{aligned}
$$

i.e., $a b c \leqslant\left(\frac{l}{3+3 \sqrt{2}}\right)^{3}$. At this point,
$$
V \leqslant \frac{1}{6}\left(\frac{l}{3+3 \sqrt{2}}\right)^{3} \text {. }
$$

Clearly, the condition for equality in the above two inequalities is $a=b=c$. Therefore, the answer is $\frac{1}{6}\left(\frac{l}{3+3 \sqrt{2}}\right)^{3}$.