Prove $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y\rfloor+\lfloor x+y\rfloor$ [duplicate]

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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.

marked as duplicate by Semiclassical, B. Goddard, Misha Lavrov, Milo Brandt, Zain PatelApr 22 '17 at 3:15

• I would say this: break $x = [x] + \{ x\}, y = [y] + \{ y\}$, and substitute in the above expression. Once you get a known expression, work backwards. – астон вілла олоф мэллбэрг Apr 22 '17 at 2:27

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 $$\lfloor2s\rfloor+\lfloor2t\rfloor\ge\lfloor2s\rfloor\ge\lfloor s+t\rfloor.$$

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 $\lfloor s + r \rfloor \le 1$

• Can you continue with your case work? I always get stuck with case work, so I need some help here. – Quantum Pizza Apr 22 '17 at 2:47
• @QuantumPizza Just work out the expressions and verify that the equation is satified.... i did case 2 as an example – Bram28 Apr 22 '17 at 2:59