Mathematical Reflections 2012, Issue 5 - Problem S241

Problem:
Let $p$ and $q$ be odd primes such that $\dfrac{p^3-q^3}{3} \geq 2pq+3.$ Prove that
$$\dfrac{p^3-q^3}{4} \geq 3pq+16.$$

Proposed by Titu Andreescu.

Solution:
Clearly, $p>q$, so $p-2 \geq q$. It must be $p-2 \neq q$, otherwise we would have $$p^3-q^3=p^3-(p-2)^3=6p^2-12p+8=6pq+8,$$ which contradicts the given inequality. So $p-4 \geq q$ and $$p^3-q^3 \geq p^3-(p-4)^3=12p^2-48p+64=12p(p-4)+64 \geq 12pq+64,$$ which gives the conclusion.