Prove that by the weighted AM-GM inequality we have
$$a^{3}+8+9(k-1) \geqslant 9 k\left(\frac{8+a^{3}}{9}\right)^{\frac{1}{4}}$$

which is equivalent to
$$\frac{a^{3}+9 k-1}{9 k} \geqslant \sqrt[n]{\frac{8+a^{3}}{9}}$$

Similarly, we can get
$$\begin{array}{l} \frac{b^{3}+9 k-1}{9 k} \geqslant\left(\frac{8+b^{3}}{9}\right)^{\frac{1}{k}} \\ \frac{c^{3}+9 k-1}{9 k} \geqslant\left(\frac{8+c^{3}}{9}\right)^{\frac{1}{k}} \end{array}$$

Let $9 k-1=m$, then we only need to prove
$$\sum \frac{a}{b^{3}+m} \geqslant \frac{3}{m+1}$$

where $m \geqslant 0$. By making the substitution $a=\frac{x}{y}, b=\frac{z}{x}, c=\frac{y}{z}$, where $x, y, z>0$, the inequality becomes
$$\sum \frac{x^{4}}{y\left(z^{3}+k x^{3}\right)} \geqslant \frac{3}{k+1}$$

By the Cauchy-Schwarz inequality, we have
$$\sum \frac{x^{4}}{y\left(z^{3}+k x^{3}\right)} \geqslant \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{\sum_{c y c} y z^{3}+k \sum_{y y c} x^{3} y}$$

Vasile's inequality tells us that
$$\begin{array}{l} \sum_{c y c} y z^{3} \leqslant \frac{1}{3}\left(x^{2}+y^{2}+z^{2}\right)^{2} \\ \sum_{c y c} x^{3} y \leqslant \frac{1}{3}\left(x^{2}+y^{2}+z^{2}\right)^{2} \end{array}$$

Thus, the proposition is proved.