Number Less One is Greater than Square Root

Lemma

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

Then for $n > 2$:

$n - 1 > \sqrt n$

Let $n \in \N$ satisfy:

$n - 1 > \sqrt n$

Then:

$\ds n - 1$

$>$

$\ds \sqrt n$

$\ds \leadstoandfrom \ \ $

$\ds \paren {n - 1}^2$

$>$

$\ds n$

squaring both sides

$\ds \leadstoandfrom \ \ $

$\ds n^2 - 3 n + 1$

$>$

$\ds 0$

moving all terms to the left hand side

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

$\ds n^2 - 3 n + 1$

$<$

$\ds \paren {n + 1}^2 - 3 \paren {n + 1} + 1$

$\ds $

$=$

$\ds n^2 - n - 1$

$\ds \leadsto \ \ $

$\ds 2 \paren {n - 1}$

$>$

$\ds 0$

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 $n \ge 3$.

$\blacksquare$