If $\displaystyle c > 0$ is an integer and not prime, then there is an integer $\displaystyle b > 0$ such that $\displaystyle b|c$ and $\displaystyle b \leq \sqrt{c}$.

Approach so far:

Contradiction, let us assume that $\displaystyle b > \sqrt{c}$.

Since $\displaystyle br = c$ for some integer $\displaystyle r$, then $\displaystyle r < c$.

I feel I can arrive at a contradiction, since $\displaystyle r \geq c$ from our initial statement, but not sure how to arrive at such.