# Prove the following floor function identity

The identity is this: $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$

I am truly stumped. Help is appreciated. Thank you!

## 2 Answers

Certainly this is true if $x$ and $y$ are integers. Suppose now $x=b_{x}+\epsilon_{x}$ and $y=b_{y}+\epsilon_{y}$ where $b_{x}$ and $b_{y}$ are integers and $0<\epsilon_{x},\epsilon_{y}<\frac{1}{2}$. Then, $$\left\lfloor 2x\right\rfloor +\left\lfloor 2y\right\rfloor =2b_{x}+2b_{y}$$ and $$\left\lfloor x\right\rfloor +\left\lfloor y\right\rfloor +\left\lfloor x+y\right\rfloor =2b_{x}+2b_{y}.$$ Now, try doing the cases (i) $\frac{1}{2}\leq\epsilon_{x},\epsilon_{y}<1$ and (ii) $0<\epsilon_{x}<\frac{1}{2}$ and $\frac{1}{2}\leq\epsilon_{y}<1$ on your own. By symmetry, (ii) handles the case (iii) $\frac{1}{2}\leq\epsilon_{x}<1$ and $0<\epsilon_{y}<\frac{1}{2}$.

Adding $1$ to $x$ or $y$ increases both sides of the stated inequality by $2$. We therefore may assume $0\leq x<1$, $0\leq y<1$, which implies $\lfloor x\rfloor=\lfloor y\rfloor=0$. We then have $$\lfloor x+y \rfloor={\bf 1}[x+y\geq1]\leq {\bf 1}\bigl[\max\{x,y\}\geq{\textstyle{1\over2}}\bigr]\leq {\bf 1}\bigl[x\geq{\textstyle{1\over2}}\bigr]+{\bf 1}\bigl[y\geq{\textstyle{1\over2}}\bigr]=\lfloor2x \rfloor+\lfloor 2y \rfloor\ .$$ Here ${\bf 1}[\ ]$ denotes the characteristic function of the indicated set.