Number Less One is Greater than Square Root

Lemma
Let $n \in \N$ be an integer.

Then for $n > 2$:


 * $n - 1 > \sqrt{n}$

Proof
Let $n \in \N$ satisfy:


 * $n - 1 > \sqrt{n}$

Then:

By definition, $n^2 - 3n + 1$ is strictly increasing when the following inequality holds:

Clearly this simplifies to:


 * $2\left({n-1}\right) > 0$

The solution to this is $n \ge 2$.

By inspection, $n = 3$ satisfies $n^2 - 3n + 1 > 0$.

Since $n^2 - 3n + 1$ is strictly increasing, the inequality holds for all $n \ge 3$.