Divisibility problem ($p \leq \sqrt{n}$)

Question

If $n \geq 2$ and $n$ is composite, then there exists a prime $p$ such that that $p \mid n$ and $p \leq \sqrt{n}$

As $n$ is composite, it follows that $n = ab$ for some $a, b \in \Bbb N$, where $1 < a < n$ and $1 < b < n$.

If $n$ was $6$, I can see how it would follow: $2|6$ and $2 \leq \sqrt{6}$.

I'm just unsure on how to prove this...

Answer 1

Hint: Suppose $a > \sqrt{n}$ and $b > \sqrt{n}$. What can you say about $ab$?