Integer as Sum of Polygonal Numbers/Lemma 2

Theorem
Let $n, m \in \R_{>0}$ such that $\dfrac n m \ge 1$.

Define $I$ to be the open real interval:


 * $I = \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$

Then:


 * For $\dfrac n m \ge 116$, the length of $I$ is greater than $4$.

Proof
We need to show that $\paren {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } - \paren {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3}} > 4$ when $\dfrac n m \ge 116$.

Let $x = \dfrac n m - 1$.

Then:

To simplify calculations, we consider:

which is true when $x \ge 115$.

Thus this condition is satisfied when $\dfrac n m \ge 116$.