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

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

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

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