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.
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 $x = 3/4$, $y = 1/4$.
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} &lt; 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] =&lt; [x+1/2] + [y+1/2]