Find $M_{min}$ if there exist constant $M$ such $f(x)If   $\dfrac{f(x)}{x^2}$monotone increasing function on $x\in (0,+\infty)$,and there exist constant $M$,such $f(x)<M,\forall x\in (0,+\infty)$,then Find the $M_{min}$
If we let $g(x)=\dfrac{f(x)}{x^2}$,then for any $x,y>0(x<y)$,we have $g(x)<g(y)$ or
$$\dfrac{f(x)}{x^2}<\dfrac{f(y)}{y^2}$$
but I don't have any idea how to start proving it,
Thanks
 A: EDIT: It is not needed to assume $f$ continuous. 
If $M_{min}>0$ then $\forall \epsilon>0: \exists x>0: f(x)>M_{min}-\epsilon$. Then for small enough $\epsilon>0$ such that $M-\epsilon>0$ we have $\exists x>0: f(x)>M-\epsilon>0$. But then
$\frac{f(x)}{x^2}<\frac{M}{x^2}\to 0^+$ when $x\to\infty$ which means that $\frac{f(x)}{x^2}$ can not be monotonically increasing. Therefore $M_{min}\leq 0$. Also if we take $f(x)=-\epsilon,\,\epsilon>0$ then $\frac{-\epsilon}{x^2}$ is monotonically increasing in $(0,\infty)$ and it follows that $M_{min}$ can not be negative (because we can take  $\epsilon$ as small as we want). So $M_{min}=0$. The last  means that from the assumptions in the question we can only infer that $M_{min}\leq 0$. But we can not infer that $M_{min}<0$ because if we assume it, then there are functions $f(x)$ for which $\frac{f(x)}{x^2}$ is monotonically increasing and $f(x)>M_{min}$.
A: The function $f$ must not have an infimum. Take $f(x) := -\frac{1}{x}$, then $f(x)$ is bounded from above by $0$ and $\frac{f(x)}{x^2} = -\frac{1}{x^3}$ is monotonically increasing, but $f(x) \to - \infty$ as $x \to 0$.
