Show that $[2x]+[2y] \geq [x]+[y]+[x+y]$ Prove that $[2x]+[2y] \geq [x]+[y]+[x+y]$ whenever $x$ and $y$ are real numbers.
The $[]$ symbol is the greatest integer or floor function.
I have proved this fact by cases, but I stumbled upon what I believe to be another way to prove the above inequality, and I was wondering if my sequence of statements are legitimate.
I make use of two lemmas that I have proved
Lemma 1. If $x$ is a real number and m is an integer, then $[x+m] = [x]+m$.
Lemma 2. $\displaystyle [x]+\left[x+\frac{1}{2} \right] = [2x]$.
My proof begins with an "obvious" statement 
$$x+y \leq x+y+1$$
I then take the floor of the inequality to get
$$[x+y] \leq [x+y]+1$$             (1)
which is true in virtue of lemma 1. 
Furthermore, if I add the following statements 
$$[x+1/2] \leq x + 1/2$$
$$[y+1/2] \leq y + 1/2$$
I procure
$$\left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2}\right] \leq x+y+1$$
which by definition of the floor function renders the equation
$$[x+y]+1 = \left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2} \right]$$        (2)
Substituting (2) for (1), I have 
$$[x+y] \leq \left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2} \right]$$
I then add $[x]$ and $[y]$ to the above inequality to produce
$$[x]+[y]+[x+y] \leq [x]+\left[x+\frac{1}{2} \right]+[y]+\left[y+\frac{1}{2} \right]$$
And in virtue of Lemma 2, the right hand side of the inequality becomes
$$[x]+[y]+[x+y] \leq [2x]+[2y].$$
I personally don't see anything wrong, except maybe for the implication made to establish (2). 
Solving these kinds of problems is solely for personal gratification, so I will greatly appreciate feedback.
Thanx.
 A: Your proof is flawed for the reason described by Adam Hughes in his comment.  It is not the case that $$\lfloor x+y \rfloor + 1 = \lfloor x + 1/2 \rfloor + \lfloor y + 1/2 \rfloor$$ from the given inequality, because simply taking the floor of the inequality $$\lfloor x + 1/2 \rfloor + \lfloor y + 1/2 \rfloor \le x + y + 1$$ does not automatically turn the inequality into an equality.  A simple counterexample is $x = y = 1/4$.  Another counterexample is $x = 3/4$, $y = 1/4$.
A: Alternative route:
Write $x=n+r$ and $y=m+s$ where $n$ and $m$ are integers and $r,s\in\left[0,1\right)$.
Then $\lfloor2x\rfloor+\lfloor2s\rfloor=2n+2m+\lfloor2r\rfloor+\lfloor2s\rfloor$
and $\lfloor x\rfloor+\lfloor y\rfloor+\lfloor x+y\rfloor=2n+2m+\lfloor r+s\rfloor$.
This shows that it is enough to prove $\lfloor2r\rfloor+\lfloor2s\rfloor\geq\lfloor r+s\rfloor$
For this discern the cases $r,s\in\left[0,0.5\right)$ and $r\in\left[0.5,1\right)\vee s\in\left[0.5,1\right)$.
A: Let $x=m+\theta_1$ and $y=n+\theta_2$ where $n,m\in\Bbb{Z}$ and $0\le\theta_1,\theta_2<1$
Case I: $0\le\theta_1<\dfrac12$ and $0\le\theta_2<\dfrac12$
Now   \begin{align}
\lfloor 2x\rfloor +\lfloor 2y\rfloor&=\lfloor 2m+2\theta_1\rfloor+\lfloor 2n+2\theta_2\rfloor\\
&=2m+2n\\
\end{align} since $0\le\theta_1<\dfrac12$ and $0\le\theta_2<\dfrac12\implies 0\le2\theta_1<1$ and $0\le2\theta_2<1$
and   \begin{align}
\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor &=\lfloor m+\theta_1\rfloor+\lfloor n+\theta_2\rfloor+\lfloor m+n+\theta_1+\theta_2\rfloor\\
&=m+n+(m+n)\\
&=2m+2n,
\end{align} since $0\le\theta_1<\dfrac12$ and $0\le\theta_2<\dfrac12\implies 0\le\theta_1+\theta_2<1$
$\color{red}{So\quad \lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor=\lfloor 2x\rfloor +\lfloor 2y\rfloor}$
Case II: $0\le\theta_1<\dfrac12$ and $\dfrac12\le\theta_2<1$
\begin{align}
\lfloor 2x\rfloor +\lfloor 2y\rfloor&=\lfloor 2m+2\theta_1\rfloor+\lfloor 2n+2\theta_2\rfloor=2m+(2n+1),
\end{align} since $0\le\theta_1<\dfrac12$ and $\dfrac12\le\theta_2<1\implies 0\le2\theta_1<1$ and $1\le2\theta_2<2$
and  since $0\le\theta_1<\dfrac12$ and $\dfrac12\le\theta_2<1\implies \dfrac12\le\theta_1+\theta_2<\dfrac32$ \begin{align}
\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor &=\lfloor m+\theta_1\rfloor+\lfloor n+\theta_2\rfloor+\lfloor m+n+\theta_1+\theta_2\rfloor\\
&=m+n+(m+n)=2m+2n, \text{if}\quad\dfrac12\le\theta_1+\theta_2<1\\ 
&=2m+2n+1, \text{if}\quad 1\le\theta_1+\theta_2<\dfrac32
\end{align}
$\color{red}{So \quad\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor\le\lfloor 2x\rfloor +\lfloor 2y\rfloor}$
Case III: $\dfrac12\le\theta_1<1$ and $0\le\theta_2<\dfrac12$. $\color{blue}{\textbf{[Same as Case II]}}$
Case IV: $\dfrac12\le\theta_1<1$ and $\dfrac12\le\theta_2<1$
\begin{align}
\lfloor 2x\rfloor +\lfloor 2y\rfloor&=\lfloor 2m+2\theta_1\rfloor+\lfloor 2n+2\theta_2\rfloor=(2m+1)+(2n+1)=2m+2n+2,
\end{align} since $\dfrac12\le\theta_1<1$ and $\dfrac12\le\theta_2<1\implies 1\le2\theta_1<2$ and $1\le2\theta_2<2$
and  since $\dfrac12\le\theta_1<1$ and $\dfrac12\le\theta_2<1\implies 1\le\theta_1+\theta_2<2$ \begin{align}
\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor &=\lfloor m+\theta_1\rfloor+\lfloor n+\theta_2\rfloor+\lfloor m+n+\theta_1+\theta_2\rfloor\\
&=m+n+(m+n+1)=2m+2n+1\\
\end{align}
$\color{red}{So \quad\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor\le\lfloor 2x\rfloor +\lfloor 2y\rfloor}$
A: If you want to prove it that way, first you need to prove
[x+y]≤[x+1/2]+[y+1/2] again as your proof is flawed.
Simply think of [x]+[y] and [x+y]
If {x}+{y} < 1 then [x] + [y] = [x+y]
If {x}+{y} >= 1 then [x] + [y] = [x+y] - 1
So, [x]+[y] is at least [x+y] - 1.
-> [x+y+1]-1 = [x+y] =< [x+1/2] + [y+1/2]
