# How to bound a ratio of integer and a real number in floor function?

I have :

$$n\le\left\lfloor \dfrac{m}{x}\right\rfloor,$$

where $n$, $m$ are positive integers and $x<1$ is a positive real number.

I would like to bound the ratio $\frac{n}{m}$. So I will get:

$$\dfrac{n}{m}\le\dfrac{1}{m}\left\lfloor \dfrac{m}{x}\right\rfloor.$$

If I use $\lfloor x\rfloor\le x$, I will get this bound:

$$\dfrac{n}{m}\le\dfrac{1}{x},$$ but this is not a tight bound for me. Can I found a better bound of $$\dfrac{1}{m}\left\lfloor \dfrac{m}{x}\right\rfloor.$$ Can I say that $$\dfrac{n}{m}\le\left\lfloor\dfrac{1}{x}\right\rfloor ?$$

$$\frac{n}m\le\left\lfloor\frac1x\right\rfloor\;;$$
for a counterexample let $n=3$, $m=2$, and $x=\frac23$. Then $3=\left\lfloor\frac2{2/3}\right\rfloor$, but $\frac32>1=\left\lfloor\frac1{2/3}\right\rfloor$.
More generally, if $m<n$ and $x=\frac{m}n$, then $n=\left\lfloor\frac{m}x\right\rfloor$, and $\frac{n}m=\frac1x$; this shows that you cannot in general hope to do better than $\frac1x$.