Show that $[2x]+[2y] \geq [x]+[y]+[x+y]$

Question

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.

Answer 1

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

Answer 2

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 $x = 3/4$, $y = 1/4$.

Answer 3

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}

$\color{red}{So \quad\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor\le\lfloor 2x\rfloor +\lfloor 2y\rfloor}$