# Inequality involving floor function

In the context of a paper I study, I want to prove an inequality. Consider $K,N$ positive integers and $a,b$ positive real numbers basically with $a=0.275$ and $b=1.1$. I try to prove the following

$$-\frac{(\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor)^2}{1-\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor/N}\frac{b\cdot\mathrm{max}\{1,N/K\}}{N}\geq-\mathrm{min}\{N,K\}\frac{ba^2}{1-a\cdot\mathrm{min}\{1,K/N\}}$$

Equivalently we have that \begin{eqnarray} -\frac{(\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor)^2}{1-\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor/N}b\cdot\mathrm{max}\{1/N,1/K\}&\geq&-\mathrm{min}\{N,K\}\frac{ba^2}{1-a\cdot\mathrm{min}\{1,K/N\}}\\ -\frac{(\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor)^2}{1-\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor/N}b\cdot\mathrm{min}\{N,K\}&\geq&-\mathrm{min}\{N,K\}\frac{ba^2}{1-a\cdot\mathrm{min}\{1,K/N\}}\\ \Leftrightarrow\frac{(\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor)^2}{1-\lfloor a\cdot\mathrm{min}\{N,K\}}/N&\leq&\frac{a^2}{1-a\cdot\mathrm{min}\{1,K/N\}} \end{eqnarray}

by exploiting $$\lfloor x\rfloor\geq x-1$$

I am trying to compare the denominators so by the above identity \begin{eqnarray} \lfloor a\cdot\mathrm{min}\{N,K\}\rfloor/N&\geq&a\cdot\mathrm{min}\{N,K\}/N-1/N\\ \Leftrightarrow-\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor/N&\leq&-a\cdot\mathrm{min}\{N,K\}/N+1/N\\ \Leftrightarrow1-\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor/N&\leq&1-a\cdot\mathrm{min}\{N,K\}/N+1/N\\ \Leftrightarrow1-\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor/N&\leq&1-a\cdot\mathrm{min}\{1,K/N\}+1/N \end{eqnarray} but I cannot reach a conclusive result here. Also I am not sure about the second step of the previous series of inequalities.

• It seems that the inequality does not hold when $N=1,K=2,a=10.1, b=11$. – mathlove Jan 15 '17 at 4:01
• @mathlove basically in my case $a=0.275$ and $b=1.1$ are fixed but I did not state that in order to simplify the expression. I thought it woulnd't matter. $N,K$ are still arbitrary positive integers, though. – mgus Jan 15 '17 at 4:06

For $0\lt a\lt 1$ and $b\gt 0$ such that $$1-\frac{\lfloor a\cdot\min\{N,K\}\rfloor}{N}\not=0\quad\text{and}\quad 1-a\cdot\min\left\{1,\frac KN\right\}\not=0$$ the inequality holds.

$$-\frac{(\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor)^2}{1-\lfloor a\cdot\mathrm{min}\{N,K\}\rfloor/N}\frac{b\cdot\mathrm{max}\{1,N/K\}}{N}\geq-\mathrm{min}\{N,K\}\frac{ba^2}{1-a\cdot\mathrm{min}\{1,K/N\}}\tag1$$

Let us separate it into two cases :

Case 1 : $K\ge N$ \begin{align}(1)&\iff -\frac{(\lfloor a\cdot N\rfloor)^2}{1-\frac{\lfloor a\cdot N\rfloor}{N}}\cdot\frac{b\cdot 1}{N}\geq-N\cdot\frac{ba^2}{1-a\cdot 1}\\\\&\iff \frac{\lfloor aN\rfloor^2}{N-\lfloor aN\rfloor}\le \frac{(aN)^2}{N-aN}\tag2\end{align}

Multiplying the both sides of $(2)$ by $(N-\lfloor aN\rfloor)(N-aN)\gt 0$ gives \begin{align}(2)&\iff (N-aN)\lfloor aN\rfloor^2\le (aN)^2(N-\lfloor aN\rfloor)\\\\&\iff a^2N^3-a^2N^2\lfloor aN\rfloor-N\lfloor aN\rfloor^2+aN\lfloor aN\rfloor^2\ge 0\\\\&\iff (aN+\lfloor aN\rfloor)(aN-\lfloor aN\rfloor)-a\lfloor aN\rfloor(aN-\lfloor aN\rfloor)\ge 0\\\\&\iff (aN-\lfloor aN\rfloor)(aN+(1-a)\lfloor aN\rfloor)\ge 0\end{align} which holds for $0\lt a\lt 1$.

Case 2 : $K\lt N$

\begin{align}(1)&\iff -\frac{(\lfloor aK\rfloor)^2}{1-\frac{\lfloor aK\rfloor}{N}}\cdot\frac{b\cdot\frac NK}{N}\geq -K\cdot\frac{ba^2}{1-a\cdot\frac KN}\\\\&\iff \frac{\lfloor aK\rfloor^2}{N-\lfloor aK\rfloor}\le \frac{(aK)^2}{N-aK}\tag3\end{align}

Multiplying the both sides of $(3)$ by $(N-\lfloor aK\rfloor)(N-aK)\gt 0$ gives $$(3)\iff (aK-\lfloor aK\rfloor)(NaK+(N-aK)\lfloor aK\rfloor)\ge 0$$ which holds for $0\lt a\lt 1$.

• Thanks, it seems that you have a lot of imagination. How did you approach this problem? Those are the kinds of things I usually encounter when I study papers. – mgus Jan 15 '17 at 6:11
• @mgus: You are welcome. Well, first, I found that separating the inequality into two cases should make it easy to deal with since the only thing that matters is the relation between $K$ and $N$. After this, I tried to simplify the inequality (hoping to factorize it) and luckily managed to factorize it. I hope this helps. – mathlove Jan 15 '17 at 6:24
• @mgus: In general, it is not very easy to deal with min, max. So, separating an (in)equality into some cases in order to eliminate min, max is a good start especially when many cases are not needed like our inequality. – mathlove Jan 15 '17 at 6:36