29. 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)
29. Let $c$ be the largest side, then $c=\sqrt{a^{2}+b^{2}}$, and
$$a^{3}+b^{3}+c^{3}=a^{3}+b^{3}+2 \sqrt{2}\left(\sqrt{\frac{a^{2}+b^{2}}{2}}\right)^{3}$$

By the weighted power mean inequality, we have
$$\begin{aligned}
\sqrt[3]{\frac{a^{3}+b^{3}+2 \sqrt{2}\left(\sqrt{\frac{a^{2}+b^{2}}{2}}\right)^{3}}{1+1+2 \sqrt{2}}} \geqslant \sqrt{\frac{a^{2}+b^{2}+2 \sqrt{2}\left(\sqrt{\frac{a^{2}+b^{2}}{2}}\right)^{2}}{1+1+2 \sqrt{2}}}= \\
\sqrt{\frac{a^{2}+b^{2}}{2}}
\end{aligned}$$

Since $\sqrt{\frac{a^{2}+b^{2}}{2}} \geqslant \frac{a+b}{2}$, we have
$$\begin{array}{c}
(\sqrt{2}+1) \sqrt{a^{2}+b^{2}} \geqslant a+b+\sqrt{a^{2}+b^{2}} \\
\sqrt{\frac{a^{2}+b^{2}}{2}} \geqslant \frac{1}{2+\sqrt{2}}\left(a+b+\sqrt{a^{2}+b^{2}}\right)=\frac{1}{2+\sqrt{2}}(a+b+c)
\end{array}$$

From (1) and (2), we get
$$a^{3}+b^{3}+c^{3} \geqslant \frac{1}{\sqrt{2}(1+\sqrt{2})^{2}}(a+b+c)^{3}$$

When $a=b=1, c=\sqrt{2}$, the equality holds. Therefore, the largest positive real number $k=\frac{1}{\sqrt{2}(1+\sqrt{2})^{2}}$.