Floors Complicated Proof

Question

Problem:

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

I have my proof down below. Its a little complicated though. This is not a duplicate of the same question on another page on math.stackexchange.com because I provide my answer.

Answer 1

Let $n=\lfloor x \rfloor$ and $m=\lfloor y \rfloor$. Consider $4$ cases:

Case 1: $x \in [n,n+\dfrac{1}{2})$ and $y \in [m,m+\dfrac{1}{2})$ Case 2: $x \in [n+\dfrac{1}{2},n+1)$ and $y \in [m,m+\dfrac{1}{2})$ Case 3: $x \in [n,n+\dfrac{1}{2})$ and $y \in [m+\dfrac{1}{2},m+1)$ Case 4: $x \in [n+\dfrac{1}{2},n+1)$ and $y \in [m+\dfrac{1}{2},m+1)$

For Case 1, we have $$\lfloor 2x \rfloor = \lfloor [2n,2n+1) \rfloor=2n$$ because $n$ is an integer.

Similarly, $$\lfloor 2y \rfloor = \lfloor [2m,2m+1) \rfloor=2m$$ because $m$ is an integer.

We also have $$\lfloor x+y \rfloor = \lfloor [n+m,n+m+1) \rfloor=n+m.$$ Now we can plug this into the original inequality: $$\lfloor 2x \rfloor + \lfloor 2y \rfloor=2n+2m$$ and $$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor=2n+2m$$ This gives $$\lfloor 2x \rfloor + \lfloor 2y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$$

For Case 2, we have $$\lfloor 2x \rfloor = \lfloor [2n+1,2n+2) \rfloor=2n+1$$ because $n$ is an integer.

As in Case 1, $$\lfloor 2y \rfloor = \lfloor [2m,2m+1) \rfloor=2m$$ because $m$ is an integer.

We also have $$\lfloor x+y \rfloor = \lfloor [n+m+\dfrac{1}{2},n+m+1\dfrac{1}{2}) \rfloor$$ This implies that $$\lfloor x+y \rfloor = n+m\;or\;n+m+1$$ Now we can plug this into the original inequality: $$\lfloor 2x \rfloor + \lfloor 2y \rfloor=2n+1+2m$$ and $$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor=2n+2m\;or\;2n+2m+1$$ This gives $$\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$$

Case 3 is similar to Case 2 by switching the $n$'s and $m$'s.

For Case 4, we have $$\lfloor 2x \rfloor = \lfloor [2n+1,2n+2) \rfloor=2n+1$$ because $n$ is an integer.

Similarly, $$\lfloor 2y \rfloor = \lfloor [2m+1,2m+2) \rfloor=2m+1$$ because $m$ is an integer.

We also have $$\lfloor x+y \rfloor = \lfloor [n+m+1,n+m+2) \rfloor=n+m+1.$$ Now we can plug this into the original inequality: $$\lfloor 2x \rfloor + \lfloor 2y \rfloor=2n+2m+2$$ and $$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor=2n+2m+1$$ This gives $$\lfloor 2x \rfloor + \lfloor 2y \rfloor > \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor.$$ Therefore, the original inequality is true!

Answer 2

There is a simple solution to this, and it involves a little casework.

Case 1: The fractional parts of $x$ and $y$ are both less than $0.5$. Then $$\lfloor 2x\rfloor=2\lfloor x\rfloor$$ $$\lfloor 2y\rfloor=2\lfloor y\rfloor$$ $$\lfloor x+y\rfloor=\lfloor x\rfloor + \lfloor y\rfloor$$ and the two sides of the inequality are equal, making it trivially true.

Case 2: The fractional parts of $x$ and $y$ are both greater than $0.5$. Then $$\lfloor 2x\rfloor=2\lfloor x\rfloor+1$$ $$\lfloor 2y\rfloor=2\lfloor y\rfloor+1$$ $$\lfloor x+y\rfloor=\lfloor x\rfloor+\lfloor y\rfloor+1$$ And the two sides of the equation look like $$2\lfloor x\rfloor+2\lfloor y\rfloor+2\ge2\lfloor x\rfloor+2\lfloor y\rfloor+1$$ Which is always true.

Case 3: The fractional part of $x$ is less than $0.5$ and the fractional part of $y$ is greater than $0.5$ (or vice versa, without loss of generality). Then $$\lfloor 2x\rfloor=2\lfloor x\rfloor$$ $$\lfloor 2y\rfloor=2\lfloor y\rfloor+1$$ And the quantity $\lfloor x+y\rfloor$ is either $0$ or $1$ more than the quantity $\lfloor x\rfloor+\lfloor y\rfloor$, and we have one of these two inequalities: $$2\lfloor x\rfloor+2\lfloor y\rfloor+2\ge2\lfloor x\rfloor+2\lfloor y\rfloor+1$$ or $$2\lfloor x\rfloor+2\lfloor y\rfloor+2\ge2\lfloor x\rfloor+2\lfloor y\rfloor$$ Both of which are always true. This is true for all cases, so it is always true.