Let $i$ and $m$ be integers such that $i \geq m \geq 2$. Suppose $k \in \mathbb{R}^{+}_{0}$, with $(m-1)\frac{1}{2}k <1$ and $m\frac{1}{2}k \geq 1$. Denote $[\frac{i}{m}]$ for the largest integer $\leq \frac{i}{m}$. How do you proof that $[\frac{i}{m}]\geq \frac{i}{2m-1}$? It's not hard to see that $\frac{i}{m}> \frac{i}{2m-1}$, but how do you continue from this point?
I already tried the following: suppose $[\frac{i}{m}]< \frac{i}{2m-1}< \frac{i}{m}$, then $\frac{i}{m}-\frac{i}{2m-1} < \frac{i}{m} - [\frac{i}{m}]<1$, but this just gives you that $i(m-1)<m(2m-1)$, which doesn't tell you anything more...
There exist integers $N,r$ such that $$i=mN+r$$ where $N\ge 1$ and $0\le r\le m-1$.
Now, we have $$\left[\frac im\right]-\frac{i}{2m-1}=N-\frac{mN+r}{2m-1}=\frac{(m-1-r)+(m-1)(N-1)}{2m-1}$$ which is non-negative since $$m-1-r\ge 0\qquad\text{and}\qquad (m-1)(N-1)\ge 0$$
It seems that the part "Suppose $k \in \mathbb{R}^{+}_{0}$, with $(m-1)\frac{1}{2}k <1$ and $m\frac{1}{2}k \geq 1$" is irrelevant.