Example 3 Let $a, b, c$ be the lengths of the three sides of a right-angled triangle, where $c$ is the length of the hypotenuse. Find the maximum value of $k$ such that $\frac{a^{3}+b^{3}+c^{3}}{a b c} \geqslant k$ holds. ${ }^{[3]}$
(Fourth Northern Mathematical Olympiad Invitational Competition)
Given $a^{2}+b^{2}=c^{2}$, let
$$
\begin{array}{l}
a=c \cos \theta, b=c \sin \theta\left(\theta \in\left(0, \frac{\pi}{2}\right)\right) . \\
\text { Then } f=\frac{a^{3}+b^{3}+c^{3}}{a b c} \\
=\frac{(\sin \theta+\cos \theta)^{3}-3 \sin \theta \cdot \cos \theta(\sin \theta+\cos \theta)+1}{\sin \theta \cdot \cos \theta} .
\end{array}
$$

Let $\sin \theta+\cos \theta=t \in(1, \sqrt{2}]$. Then
$$
f=\frac{-t^{2}+t+2}{t-1}=-(t-1)+\frac{2}{t-1}-1 \text {, }
$$

where $t \in(1, \sqrt{2}]$.
By monotonicity, it is easy to see that $f_{\text {min }}=f(\sqrt{2})=2+\sqrt{2}$.
At this point, $a=b$.
In conclusion, $k_{\max }=2+\sqrt{2}$.