Solution:
Observe that $2x^{2} + 16 = 2(x^{2} + 8) = 2(x + 2)(x^{3} = 2x + 4)$. From AM-GM:
$$\sqrt{2x^{3} + 16} = \sqrt{(2x + 4)(x^{3} - 2x + 4)} \leq \frac{2x + 4 + x^{3} - 2x + 4}{2} = \frac{x^{2} + 8}{2}$$
By adding the inequality (1) obtained for $x = a$, $x = b$ and $x = c$ it suffices to prove:
$$a^{3} b^{3} + 8a^{3} + b^{3} c^{3} + 8b^{3} + c^{3} c^{3} + 8c^{3} \leq 2 \cdot 24^{3},$$
Since $a^{3} b^{3} + b^{3} c^{3} + c^{3} a^{3} \leq \frac{\left(a^{3} + b^{3} + c^{3}\right)^{3}}{3}$, using $a^{3} + b^{2} + c^{2} = 48$, we get the stated inequality.

Equality holds only when $a = b = c = 4$.