Showing that $f(x)/x$ is decreasing on $[0, \infty) $ I want to prove that if $f:[0, \infty) \to [ 0, \infty )$,$f(0)=0$ and $f(x)>0$ for all $x>0$, $f$ is increasing and $f'$ is decreasing then $f(x)/x$ is decreasing for $x>0$.
Here's my attempt:
Let $x,y \in (0, \infty)$ and suppose that $x\le y$. By the mean value theorem, we have that $f(x)/x=f'(c)$ and $f(y)/y=f'(d)$ for some $c \in (0, \infty)$ and some $d \in (0, \infty)$. If we show that $c\le d$, we'll be done because then $f'(d)\le f'(c)$, i.e., $f(y)/y \le f(x) /x$. So assume that $d<c$. Then $f'(c) \le f'(d)$.
I haven't been able to proceed from here. Hints would be appreciated.

$f(x)=\int_0^{x} f'(t)dt \geq \int_0^{x} f'(x)dt =xf'(x)$. Hence $\frac d {dx} (\frac {f(x)}x )=\frac {xf'(x)-f(x)} x \leq 0$ and this implies that $\frac {f(x)}x$ is decreasing.
