<s>[INST] Floor function inequality I am racking my brain trying to get this problem solved and I can't seem to break it...
Let $m, n$ be positive integers, with $m > 1$.  Prove
$$\left\lfloor\frac{n}m\right\rfloor+\left\lfloor\frac{n+1}m\right\rfloor\le\left\lfloor\frac{2n}m\right\rfloor$$
I started trying to use the inequalities that say 
$$\lfloor x\rfloor + \lfloor y\rfloor + \lfloor x+y\rfloor \le \lfloor 2x\rfloor + \lfloor 2y\rfloor\;.$$
From there I rewrote both sides with the notion that for a real number $y$, 
$y = \lfloor y\rfloor + \{y\}$, where $\{y\}$ is the fractional part, but I keep going in circles and can not get to the desired result...
 [/INST] Let $a={\sf floor}(\frac{n}{m})$ and $b={\sf floor}(\frac{n+1}{m})$.
We have $a \leq \frac{n}{m}$ and $b \leq \frac{n+1}{m}$, so $a+b \leq \frac{2n+1}{m}$. If this were an equality, it would imply that both $a =\frac{n}{m}$ and $b = \frac{n+1}{m}$, so that $m$ would divide both $n$ and $n+1$, and hence divide $n+1-n=1$ also, which is absurd.
So we have the strict inequality $a+b \lt \frac{2n+1}{m}$, yielding $m(a+b) \lt 2n+1$. Since $m,a+b$ and $2n+1$ are all integers, we deduce $m(a+b) \leq 2n$. So $a+b$ is an integer lower than $\frac{2n}{m}$ ; it is therefore smaller than the floor of  $\frac{2n}{m}$ as wished.
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