<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] If you want to use you that inequality of yours let $x=\dfrac{n}{m}$ and $y=\dfrac{1}{m}:$
$$\bigg\lfloor \frac{n}{m}\bigg\rfloor+\bigg\lfloor \frac{1}{m}\bigg\rfloor+\bigg\lfloor \frac{n+1}{m}\bigg\rfloor\leq\bigg\lfloor \frac{2n}{m}\bigg\rfloor+\bigg\lfloor \frac{2}{m}\bigg\rfloor$$
Since $m>1\Rightarrow \bigg\lfloor\dfrac{1}{m}\bigg\rfloor=0$. If additionally $m>2$ then also $\bigg\lfloor\dfrac{2}{m}\bigg\rfloor=0$ so all we need to do is check for $m=2:$
$$\bigg\lfloor \frac{n}{2}\bigg\rfloor+\bigg\lfloor \frac{n+1}{2}\bigg\rfloor\leq n+1$$
Since $\displaystyle\bigg\lfloor \frac{n}{2}\bigg\rfloor+\bigg\lfloor \frac{n+1}{2}\bigg\rfloor\leq \dfrac{n}{2}+\dfrac{n+1}{2}=n+\dfrac{1}{2}<n+1$ this case is verified.
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