By the AM-GM inequality we have
$$8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}} \le 9\sqrt[3]{\frac{8abc + \frac{a^3+b^3+c^3}{3}}{9}} = 3\sqrt[3]{a^3 + b^3 + c^3 + 24abc}$$
Hence it suffices to prove that
$$3(a + b + c) \ge 3\sqrt[3]{a^3 + b^3 + c^3 + 24abc}$$
or
$$(a + b + c)^3 \ge a^3 + b^3 + c^3 + 24abc.$$
Expanding the left-hand side, this is equivalent to
$$a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 \ge 6abc,$$
and the above inequality is easily proved by the AM-GM inequality. Equality holds if
$$abc = \frac{a^3 + b^3 + c^3}{3} \quad \text{and} \quad a^2b = ab^2 = b^2c = bc^2 = c^2a = ca^2$$
that is, when $a = b = c$.