# Proof of $\lfloor 6n \rfloor + \lfloor n \rfloor \ge 2\lfloor 2n \rfloor + \lfloor 3n \rfloor$.

I used this proof to this : $\displaystyle \frac{6n!n!}{2n!2n!3n!} \in \mathbb{Z}$.

My proof:

$$\displaystyle \lfloor x+y\rfloor \ge \lfloor x \rfloor + \lfloor y \rfloor$$

$$\lfloor 6x \rfloor \ge 2\lfloor3x\rfloor$$, so $$\lfloor 6x \rfloor + \lfloor x\rfloor \ge 2\lfloor3x\rfloor + \lfloor x \rfloor =\lfloor3x\rfloor +\lfloor x \rfloor +\lfloor 4x - x\rfloor \ge \lfloor3x \rfloor + \lfloor x\rfloor +\lfloor4x\rfloor -\lfloor x\rfloor \ge \lfloor3x\rfloor + \lfloor 4x\rfloor \ge \lfloor3x \rfloor + 2\lfloor2x\rfloor$$

But I don't know about this : $\lfloor4x - x\rfloor \ge \lfloor4x \rfloor - \lfloor x\rfloor.$ Am I right?

• Try $x=\frac14$ – Hagen von Eitzen Oct 6 '16 at 17:02
• @HagenvonEitzen thanks a lot. – openspace Oct 6 '16 at 17:09
• For what it is worth, $\lfloor 6x\rfloor$ entirely determines $\lfloor 3x\rfloor, \lfloor 2x\rfloor,$ and $\lfloor x\rfloor$, and the different can be evaluated on a case-by-case, based on $\lfloor 6x\rfloor\bmod 6$. So there should be a brute force proof. – Thomas Andrews Oct 6 '16 at 17:16
• @HagenvonEitzen is it possible to prove it using my idea? Or should I consider some intervals : $[0..1)$ and $[1..\infty]$? – openspace Oct 6 '16 at 17:17
• It probably isn't going to work that way, @openspace – Thomas Andrews Oct 6 '16 at 17:18