Is $a^p/p+b^q/q \ge ab$ true of any rational $p$, $q$ greater than $2$?

Question

From Beckingbach & Bellman, Introduction to Inequalities, p.61, they prove that

(1) $a^p/p+b^q/q \ge ab$

where $a$ and $b$ are non-negative integers and $p$ and $q$ are rational numbers, and

$1/p+1/q=1$ .

Can equation (1) be proven or disproven for the case of all rational $p$ and $q$ without the further condition, and $p>2$, $q>2$?

Answer 1

It doesn't hold for p=q=1 obviously for a=b=3.