Show that if $f'$ is strictly increasing, then $\frac{f(x)}{x}$ is increasing over $(0,\infty)$
Solution 1:
By MVT, for some $c \in (0,x)$ such that $$\frac{f(x)}{x}=\frac{f(x)-f(0)}{x-0}=f'(c)$$
Since $x>c$, this gives us that $$f'(x)>f'(c)=\frac{f(x)}{x}$$
Thus, $f'(x)x-f(x)>0$ if $x>0$.
But the derivative of $\frac{f(x)}{x}$ is $$\frac{f'(x)x-f(x)}{x^2}>0$$Our proof is done.