I have function $$D(M\gamma)=\frac{M\gamma K}{K+\gamma(M\gamma-1)}\tag{1}$$ where $K$ is some positive constant. In this case, how to show that $$D(M(\gamma-1))<D(M\gamma),~~\text{for } \gamma<\bigg\lceil\sqrt{\frac{K}{M}}\bigg\rceil.\tag{2}$$

My Try:

I tried to show that $$D(M\gamma)-D(M(\gamma-1))>0$$ which after some manipulations results in following inequality $$\sqrt{\frac{K}{M}+1}>\gamma$$ But I think this does not mean that $$D(M(\gamma-1))<D(M\gamma),~~\text{for } \gamma<\bigg\lceil\sqrt{\frac{K}{M}}\bigg\rceil$$ because $$\sqrt{x+1}-\lceil\sqrt{x}\rceil$$ is not always greater than zero. So, How to prove the inequality in (2) for the function presented in (1). Any help in this regard will be much appreciated. Thanks in advance.

  • $\begingroup$ But $\frac{M\gamma K}{K+\gamma(M\gamma-1)}$ is not a function of $M\gamma$... $\endgroup$ – Fabio Lucchini Sep 12 '18 at 9:42
  • $\begingroup$ then whose function is it? $\endgroup$ – Frank Moses Sep 12 '18 at 9:44
  • $\begingroup$ @FabioLucchini as I am assuming $M$ to be constant. Then $\gamma$ is a variable and $D(M\gamma)$ is a function of $\gamma$ $\endgroup$ – Frank Moses Sep 12 '18 at 9:46
  • $\begingroup$ You can consider as function of $\gamma$ but not of $M\gamma$. $\endgroup$ – Fabio Lucchini Sep 12 '18 at 9:54
  • 1
    $\begingroup$ @FabioLucchini actually the function that is to be maximized is as follows $$D(n)=\frac{nK}{K+\bigg\lceil\frac{n}{M}\bigg\rceil(n-1)}.$$ Now if we put $n=M\gamma$ with $\gamma$ an integer we get $$D(M\gamma)=\frac{M\gamma K}{K+\gamma(M\gamma-1)}$$ $\endgroup$ – Frank Moses Sep 12 '18 at 23:45

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