If $f'$ is increasing and $f(0)=0$, then $f(x)/x$ is increasing Let $a>0$ and $f:[0,a] \to \mathbb{R}$ continuous function that is twice differentiable on $(0,a).$ Also $f(0)=0$ and $f'$ is strictly increasing function on $(0,a).$ 
I have to show that the function $g$ defined as $$g(x) = \frac{f(x)}{x}$$ is strictly increasing on $(0,a].$

Progress: I computed $g'$ and got $g'(x) = \frac{f'(x)x-f(x)}{x^2}.$ I have to show that $g'(x)>0 \ \ \forall x \in (0,a].$  Since $x^2$ is always positive for $x \in (0,a]$ I have to show that $f'(x)x-f(x) >0 \ \ \forall x \in (0,a].$ I know both terms are positive, but I don't know how to show that $f'(x)x>f(x) \ \ \forall x \in (0,a].$
 A: By Mean Value Theorem we have for $0 < x < a$ $$\frac{f(x)}{x} = \frac{f(x) - f(0)}{x - 0} = f'(\xi) < f'(x)$$ because $f'$ is strictly increasing and $0 < \xi < x$. Thus $xf'(x) - f(x) > 0$ for all $x \in (0, a)$. And therefore $$g'(x) = \frac{xf'(x) - f(x)}{x^{2}}> 0$$ and hence $g(x) = f(x)/x$ is strictly increasing in $(0, a]$. There is no need to assume existence of $f''(x)$.
A: Observe that $f(x)=\int_0^x f'(t)\,dt$. This integral is less than $\int_0^x f'(x)\,dt=xf'(x)$ because $f'$ is strictly increasing.
A: If we are allowed to assume that $f'$ is continuous on $[0,a]$  we can write
$$g(x)=\int_0^1 f'(\tau x)\>d\tau\qquad(x\geq0)\ ,$$
which is clearly increasing as a function of $x$.
A: As $f'$ is strictly increasing, we know that $f''(x)\ge 0$ for all $x\in (0,1)$.
In order to show that $g(x)$ is increasing, we show that $g'(x)>0$ for all $x\in (0,a)$. As $g'(x)=\frac{f'(x)x-f(x)}{x^2}$ and $x^2>0$ it suffices to show that $h(x):=f'(x)x-f(x)>0$ for $0<x<a$. 
Note that $h'(x)=f''(x)x\ge 0$ so that $h$ is non-decreasing.
Especially, $h(x)\ge 0$ for all $x\in (0,a)$.
Also $h(0)=0$ so that if $h(x)=0$ for some $x\in(0,a)$ then $h(t)=0$ for all $t\in(0,x)$, hence $h'(t)=0$ for all $t\in(0,x)$, hence $f''(t)=0$ for all $t\in (0,x)$, hence $f'(x)=f'(0)$, contradicting the given fact that $f'$ is strictly increasing. We conclude that $h(x)>0$ for $0<x<a$, as was to be shown.
A: Let $A= [x,y]\subset(0,a]$ with $x<y$.Then $f$ and $g$ are contiuous on $A$ and differentiable in $(x,y)$, so we may apply the $MVT$ to find $x<c<y$ such that $$c^{-1}f'(c)-c^{-2}f(c)=c^{-1}(f'(c)-g(c))=\frac{g(y)-g(x)}{y-x}$$ from which we obtain $$(y-x)c^{-1}(f'(c)-g(c))=g(y)-g(x)$$ so we only need to show that $$f'(c)-g(c)>0$$ or $$cf'(c)-f(c)>0$$ So let's consider $h(x)=xf'(x)-f(x)$. 
Now, $h'(x)=f'(x)+xf''(x)-f'(x)=xf''(x)$ so $h\geq 0$ on $[0,a]$ and $h(x)=0$. 
But if there is a $y\in (0,a)$ such that $h(y)=0$, we can apply Rolle's Theorem to say that there is a $0<b<y<a$ such that $h'(b)=bf''(b)=0)\Rightarrow f''(b)=0$ which says $f'$ is not strictly increasing on $0,a)$. Therefore $h>0$ on $(0,a)$, which is what we wanted. 
A: Hint: The function $g$ is strictly increasing because $f$ is strictly convex. Since $f'>0$, then...
