Prove $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y\rfloor+\lfloor x+y\rfloor$

Question

Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.

If anybody could post a simple solution (no complicated abstract theories or calc :D) to the above question, I would greatly appreciate it. Thanks!

Also, if possible, the solution shouldn't wander too much away from floor, ceiling, fraction functions, and other related functions.

Answer 1

Let $X=m+s$ and $y=n+t$ where $m,n\in\mathbb Z$ and $0\le s,t\lt1.$ Then

$$\lfloor2x\rfloor+\lfloor 2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor x+y\rfloor$$ $$\iff\lfloor2m+2s\rfloor+\lfloor2n+2t\rfloor\ge\lfloor m+s\rfloor+\lfloor n+t\rfloor+\lfloor m+n+s+t\rfloor$$ $$\iff2m+\lfloor2s\rfloor+2n+\lfloor2t\rfloor\ge m+\lfloor s\rfloor+n+\lfloor t\rfloor+m+n+\lfloor s+t\rfloor$$ $$\iff\lfloor2s\rfloor+\lfloor2t\rfloor\ge\lfloor s+t\rfloor.$$ Without loss of generality, suppose $s\ge t.$ Then $2s\ge s+t,$ so $$\lfloor2s\rfloor+\lfloor2t\rfloor\ge\lfloor2s\rfloor\ge\lfloor s+t\rfloor.$$

Answer 2

HINT

Let $x = n + r$ with $n$ an integer and $0 \le r <1$

Similarly, $y = m + s$ with $m$ an integer and $0 \le s < 1$

Now there are 4 cases to consider:

1. $0 \le r < \frac{1}{2}$ and $0 \le s < \frac{1}{2}$

2. $0 \le r < \frac{1}{2}$ and $\frac{1}{2} \le s <1$

3. $\frac{1}{2} \le r < 1 $ and $0 \le s < \frac{1}{2}$

4. $\frac{1}{2} \le r < 1$ and $\frac{1}{2} \le s <1$

Just to show case 2:

$\lfloor 2x \rfloor + \lfloor 2y \rfloor = \lfloor 2n + 2r \rfloor + \lfloor 2m + 2s \rfloor = 2n + 0 + 2m + 1$

$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x + y\rfloor = n + m + n + m + \lfloor s +r \rfloor$ where $\lfloor s + r \rfloor \le 1$