Proof that $ [2x] + [2y] \ge [x]+ [x+y] + [y]$?

Question

I have to proof (or disprove) the following

$ [2x] + [2y] \ge [x]+ [x+y] + [y]$ for $x,y \in \mathbb{R}$. [x] and [y] means the floor-function.

Can I do the following?

(1) Assume $x,y \in [0,1]$, then [x] = [y] = 0 and $0 \le [x+y] \le 1$.

(2) First case:$ [x+y] = 0$, then $[2x] + [2y] \le [x] + [y]$

(3) Second case:$ [x+y] = 1$, then x,y or both x and y are greater than 1/2, which means: $[2x]+ [2y] \ge 1 = [x] + [x+y] + [y]$.

Thank you.

Answer 1

Hint:

• Assume that $x,y \in [0,1)$.
• If $\lfloor x + y \rfloor = 1$ then $x \geq \frac{1}{2}$ or $y \geq \frac{1}{2}$.

I hope this helps $\ddot\smile$