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.
Best Answer
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 $r\in\left[0.5,1\right)\vee s\in\left[0.5,1\right)$.