# How is this proposition true?

I have a function of $n$, where $n$ is an integer, defined as follows $$D(n)=\frac{nK}{K+\bigg\lceil\frac{n}{M}\bigg\rceil(n-1)},$$ where $K$ are $M$ are some constant positive integers. For this function how we can say that $$D(M(\gamma-1)+\beta)<D(M\gamma),~~~\text{for }0<\beta<M,\gamma\in \mathbb{N},\beta \in \mathbb{N}.$$ I saw this inequality in a research paper but my verification does not show that the above inequality is right. Any help in proving the above inequality will be much appreciated. Thanks in advance.

My Try:

I tried to show that $$D(M\gamma)-D(M(\gamma-1)+\beta)>0.$$ But this inequality leads to $$(K-\gamma)(1-\frac{\beta}{M})>0.$$ Which, I think, is only true when $K>\gamma$ and not for all values of integer $\gamma$. I saw

• Should the statement be interpreted as "for all $\gamma$", or as "there is a $\gamma$" ? – Yves Daoust Sep 13 '18 at 15:24

I tried to show that $D(M\gamma)-D(M(\gamma-1)+\beta)>0.$ But this inequality leads to $(K-\gamma)(1-\frac{\beta}{M})>0.$ Which, I think, is only true when $K>\gamma$ and not for all values of integer $\gamma$.

I think you are right.

What I've got is as follows :

\begin{align}&D(M\gamma)-D(M(\gamma-1)+\beta) \\\\&=\frac{M\gamma K}{K+\bigg\lceil\frac{M\gamma}{M}\bigg\rceil(M\gamma-1)}-\frac{(M\gamma-M+\beta)K}{K+\bigg\lceil\frac{M\gamma-M+\beta}{M}\bigg\rceil(M\gamma-M+\beta-1)} \\\\&=\frac{M\gamma K}{K+\gamma (M\gamma-1)}-\frac{(M\gamma-M+\beta)K}{K+\gamma(M\gamma-M+\beta-1)} \\\\&=\frac{M\gamma (K+\gamma(M\gamma-M+\beta-1))-(M\gamma-M+\beta)(K+\gamma (M\gamma-1))}{(K+\gamma (M\gamma-1))(K+\gamma(M\gamma-M+\beta-1))}K \\\\&=\frac{(M-\beta)(K-\gamma)}{(K+\gamma (M\gamma-1))(K+\gamma(M\gamma-M+\beta-1))}K \end{align}

So, we see that $$D(M\gamma)\gt D(M(\gamma-1)+\beta)\iff K\gt \gamma$$

Therefore, $$D(M(\gamma-1)+\beta)<D(M\gamma)$$ is not always true.

A counter example is $$K=1,\quad M=2,\quad \gamma=1,\quad \beta=1$$ for which we have $$D(M(\gamma-1)+\beta)=D(1)=1\color{red}{=} 1=D(2)=D(M\gamma)$$