A quick question on inequalities with floor function. For any $x\in\mathbb{R}$, denote $\lfloor x\rfloor:=\max\{n\in\mathbb{Z}\mid n\leq x\}$, i.e. the floor function. 
Show that for any $x\in\mathbb{R}$ and $m,n\in\mathbb{N}$ with $m\leq n$
$$\frac{\lfloor nx\rfloor+1}{n}\leq\frac{\lfloor mx\rfloor+1}{m}.$$
I tried to show this directly and by contradiction but seem to get nowhere, so any help is greatly needed. Thanks in advance
 A: Take $x=1.1$, $m=9$, $n=10$.
Then, we have that
$$\frac{\lfloor11\rfloor+1}{10}=1.2>\frac{\lfloor9.9\rfloor+1}{9}=1.111\ldots$$
In other words, it is false.
You can find more counterexamples choosing $x$, $m$ and $n=m+1$ such that $mx$ has a "big" (next to $1$) fractional part and $nx$ has it small.
A: The claim in the problem is false as shown by the counter-example Rolf Hoyer provided in the comments. However, the following claim is true:

Claim:
  $$\frac{\lfloor mx\rfloor+1}{n}\leq\frac{\lfloor nx\rfloor+1}{m}$$

Proof:
$$\forall~m,n\in\Bbb{N}~\mid m\leq n\implies mx\leq nx~\forall~x\in\Bbb{R}$$
$$mx\leq nx\implies \lfloor mx\rfloor\leq \lfloor nx\rfloor\implies \lfloor mx\rfloor+1\leq \lfloor nx\rfloor+1\tag1$$
Now, we also have,
$$m\leq n\iff \frac{1}{m}\geq \frac{1}{n}\iff \frac{1}{n}\leq \frac{1}{m}\tag2$$
Now, we use the following identity on $(1)$ and $(2)$:

$$a\leq b~\land~c\leq d\implies ac\leq bd$$

Using that we get,
$$\frac{\lfloor mx\rfloor+1}{n}\leq\frac{\lfloor nx\rfloor+1}{m}$$
