Proving $\frac{n}{r+\delta_1-1}\lfloor\frac{r+\delta_1-2}{r}\rfloor-\frac{n}{r+\delta_2-1}\lfloor\frac{r+\delta_2-2}{r}\rfloor\geq0$, with conditions

Question

Given $n,r,\delta_1,\delta_2$ with $n > r$, $\delta_1 > \delta_2$, both $\delta_1,\delta_2$ are $\geq 2$ and both $r+\delta_1-1, r+\delta_2-1$ divides $n$. I want to prove that \begin{equation*} \frac{n}{r+\delta_1-1} \cdot \left\lfloor\dfrac{r+\delta_1-2}{r}\right\rfloor - \frac{n}{r+\delta_2-1}\cdot \left\lfloor\dfrac{r+\delta_2-2}{r}\right\rfloor \geq 0. \end{equation*}

I am trying to prove it by arguing that

\begin{equation*} \dfrac{\left\lfloor\dfrac{r+\delta_1-2}{r}\right\rfloor}{\left\lfloor\dfrac{r+\delta_2-2}{r}\right\rfloor} \geq \dfrac{r+\delta_1-1}{r+\delta_2-1} \end{equation*}

But I am unable to do that. Any help is appreciated.

Answer 1

This is false: Take $$n=12,r=2,\delta_1=3,\delta_2=2,$$ which satisfy the conditions, but $$\frac{12}4\left\lfloor\frac 3 2\right\rfloor-\frac{12}3\left\lfloor\frac 2 2\right\rfloor=3-4<0.$$