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.
 A: 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!
A: 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.
