Given $p,q$ with $p>1,q\ge\frac1 p$, can I find $c>0$ s.t. $pqa^{pq}\ge ca^q$ for all $a\in(0,1]$?

Question

Let $p>1$ and $q\ge\frac1p$. Can I show that there is a $c>0$ s.t. $pqa^{pq}\ge ca^q$ for all $a\in (0,1]$?

Clearly, for all $r\ge0$, it holds $a^r\in(0,1]$. And for all $r\ge1$, it holds $a^r\le a$. So, I obtain $a^{pq}\le a^q$. But that's not helpful.

Answer 1

$pqa^{pq} \geq ca^q \Leftrightarrow a^{(p-1)q}\geq c/(pq)$. Now let $a \rightarrow 0$…