If $f'$ is increasing and $f(0)=0$, then $f(x)/x$ is increasing Let $a&gt;0$ 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 $g'(x)&gt;0 \ \ \forall x \in (0,a].$  Since $x^2$ is always positive for $x \in (0,a]$ I have to show that $f'(x)x-f(x) &gt;0 \ \ \forall x \in (0,a].$ I know both terms are positive, but I don't know how to show that $f'(x)x&gt;f(x) \ \ \forall x \in (0,a].$
By Mean Value Theorem we have for $0 &lt; x &lt; a$ $$\frac{f(x)}{x} = \frac{f(x) - f(0)}{x - 0} = f'(\xi) &lt; f'(x)$$ because $f'$ is strictly increasing and $0 &lt; \xi &lt; x$. Thus $xf'(x) - f(x) &gt; 0$ for all $x \in (0, a)$. And therefore $$g'(x) = \frac{xf'(x) - f(x)}{x^{2}}&gt; 0$$ and hence $g(x) = f(x)/x$ is strictly increasing in $(0, a]$. There is no need to assume existence of $f''(x)$.
Hint: The function $g$ is strictly increasing because $f$ is strictly convex. Since $f'&gt;0$, then...