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!
 A: 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 $0<\epsilon_{y}<\frac{1}{2}$.
A: 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.
