Problem 1420

28. From the generalization of Cauchy's inequality, we have
$$\begin{array}{l} \left(a b c+a b c+\cdots+a b c+\frac{a^{3}+b^{3}+c^{3}}{3}\right) \cdot \\ (1+1+\cdots+1+1)(1+1+\cdots+1+1) \geqslant \\ \left(\sqrt[3]{a b c}+\sqrt[3]{a b c}+\cdots+\sqrt[3]{a b c}+\sqrt[3]{\frac{a^{3}+b^{3}+c^{3}}{3}}\right)^{3} \end{array}$$

That is,
$$81\left(8 a b c+\frac{a^{3}+b^{3}+c^{3}}{3}\right) \geqslant\left(8 \sqrt[3]{a b c}+\sqrt[3]{\frac{a^{3}+b^{3}+c^{3}}{3}}\right)^{3}$$

We now prove that $[3(a+b+c)]^{3} \geqslant 81\left(8 a b c+\frac{a^{3}+b^{3}+c^{3}}{3}\right)$, which is equivalent to
$$(a+b+c)^{3} \geqslant 24 a b c+a^{3}+b^{3}+c^{3}$$

This inequality is equivalent to $a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^{2}+b^{2}\right) \geqslant 6 a b c$, which can be easily obtained by the AM-GM inequality. $\square$

Therefore,
$$3(a+b+c) \geqslant 8 \sqrt[3]{a b c}+\sqrt[3]{\frac{a^{3}+b^{3}+c^{3}}{3}}$$