If I'm asked to prove that $n \le m$, is it sufficient to show that $n < m$? I have a homework question, which is to prove by induction that
$\sum\limits_{r=1}^{n} \frac{1}{\sqrt{r}} \leq 2\sqrt{n}$ for every integer $n \geq 1$.
I've managed to show by induction that 
$\sum\limits_{r=1}^{k+1} \frac{1}{\sqrt{r}} \leq 2\sqrt{k} + \frac{1}{\sqrt{k}}$
and since $2\sqrt{k} + \frac{1}{\sqrt{k}} \lt 2\sqrt{k+1}$, 
$\sum\limits_{r=1}^{k+1} \frac{1}{\sqrt{r}} \lt 2\sqrt{k+1}$, given that the conjecture is true for $n = k$.
But is this enough to prove the original statement, since I only attempted to show that the sum is strictly smaller than the right-hand side? I feel like there's a nicer way to prove this.

Yes! If you have proved $n<m$, then you have proved both $n\leq m$ and $n\neq m$.
