Seven, (25 points) Let $a$, $b$, and $c$ be the lengths of the three sides of a right 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. (Provided by Li Tiehan)
$$
\text { VII. } k=\left(\frac{a^{3}+b^{3}+c^{3}}{a b c}\right)_{\min } \text {. }
$$

Let $y=\frac{a^{3}+b^{3}+c^{3}}{a b c}$, and set
$$
a=c \cos \theta, b=c \sin \theta, \theta \in\left(0, \frac{\pi}{2}\right) .
$$

Then $y=\frac{\sin ^{3} \theta+\cos ^{3} \theta+1}{\sin \theta \cdot \cos \theta}$
$$
\begin{array}{l}
=\frac{(\sin \theta+\cos \theta)(1-\sin \theta \cdot \cos \theta)+1}{\sin \theta \cdot \cos \theta} \\
=\frac{\sin \theta+\cos \theta+1}{\sin \theta \cdot \cos \theta}-(\sin \theta+\cos \theta) .
\end{array}
$$

Let $\sin \theta+\cos \theta=t$. Then
$$
\sin \theta \cdot \cos \theta=\frac{t^{2}-1}{2} \text {. }
$$

Thus, $y=\frac{t+1}{\frac{t^{2}-1}{2}}-t=\frac{2}{t-1}-t$.
Note that $t=\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right) \in(1, \sqrt{2}]$, the function $y$ is decreasing on the interval $(1, \sqrt{2}]$, hence
$$
y \geqslant \frac{2}{\sqrt{2}-1}-\sqrt{2}=2+\sqrt{2} \text {. }
$$

Therefore, $k=2+\sqrt{2}$.