<s>[INST] Floors Complicated Proof 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.
 [/INST] 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.
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