Inequality with floored squareroots

Question

$(\lfloor \sqrt{n}\rfloor +1)^2\ge n+1$, for all $n\in \mathbb{N}$.

I have convinced myself that this is true, but would like to see a formal proof.

Answer 1

Let $\lfloor \sqrt{n}\rfloor=k$.

Note that with $k \le \sqrt{n} <k+1$, squaring each side would give that $$k^2 \le n < k^2+2k+1 \Leftrightarrow k^2 \le {n} \le k^2+2k $$ Since $\lfloor \sqrt{n}\rfloor=k$, We have that $$n+1 \le k^2+2k+1=(\lfloor \sqrt{n}\rfloor+1)^2$$

Answer 2

This can be seen without any computation. Suppose otherwise that $(\lfloor \sqrt n \rfloor+1)^2 < n+1$ for some $n \in \mathbb N$. Then since the LHS is an integer, $(\lfloor \sqrt n \rfloor+1)^2 \le n$, and so $\lfloor \sqrt n \rfloor+1 \le \sqrt{n}$.

This directly contradicts the definition of floor: $\lfloor \sqrt n \rfloor$ should be the largest integer $\le \sqrt{n}$, but here is a larger integer satisfying the same bound. The contradiction proves the desired inequality.