42. In the right-angled $\triangle A B C$, find the largest positive real number $k$ such that the inequality $a^{3}+b^{3}+c^{3} \geqslant k(a+$ $b+c)^{3}$ holds. (2006 Iran Mathematical Olympiad)
42. $a^{3}+b^{3}+c^{3} \geqslant k(a+b+c)^{3} \Leftrightarrow k \leqslant\left(\frac{a}{a+b+c}\right)^{3}+\left(\frac{b}{a+b+c}\right)^{3}+$ $\left(\frac{c}{a+b+c}\right)^{3}$

Let $x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$, then because $a^{2}+b^{2}=c^{2}$, so $x^{2}+y^{2}=z^{2}$
Also, because $x+y+z=1$, so
$$x^{2}+2 x y+y^{2}=1-2 z+z^{2}$$
Subtracting (1) from (2) gives $x y=\frac{1-2 z}{2}$.

Therefore,
$$\begin{aligned}
x^{3}+y^{3}+z^{3}= & (x+y)\left(x^{2}+y^{2}-x y\right)+z^{3}= \\
& (x+y)\left(z^{2}-x y\right)+z^{3}= \\
& -x y(x+y)+z^{2}(x+y+z)= \\
& -x y(x+y)+z^{2}= \\
& -\frac{(1-2 z)(1-z)}{2}+z^{2}=\frac{3 z-1}{2}
\end{aligned}$$

Since $z^{2}=x^{2}+y^{2} \geqslant 2 x y=1-2 z$, then noting that $z>0, z^{2}+2 z-1 \geqslant 0$, solving gives $z \geqslant$ $\sqrt{2}-1$

Therefore, $x^{3}+y^{3}+z^{3} \geqslant \frac{3(\sqrt{2}-1)-1}{2}=\frac{3 \sqrt{2}-4}{2}$. Hence $k \leqslant \frac{3 \sqrt{2}-4}{2}$.
When $a=b=1, c=\sqrt{2}$, $k=\frac{3 \sqrt{2}-4}{2}$ holds. Therefore, $k_{\text {max }}=\frac{3 \sqrt{2}-4}{2}$.