A secant inequality for convex functions Suppose $f(0) =0 $ and  $0<f''(x)<\infty (\forall$  $x>0)$, then $\frac{f(x)}{x}$ strictly increases as $x$ increases.  
I have shown that $f'(x)-\frac{f(x)}{x}  = \frac{1}{2}xf''(c)$, for some $c\in (0,x)$. How do I proceed from here? 
 A: The hypothesis $f'' \geq 0$ means that your function is convex.  There are many standard inequalities involving convex functions.  This is a special case of one of them, which I call the two secant inequality:

If $f$ is convex and $a < x < b$, $\frac{f(x)-f(a)}{x-a} \leq \frac{ f(b)-f(a)}{b-a}$.

(Since your function satisfies $f'' > 0$ it is strictly convex, and then this and other related inequalities can be taken to be strict.)
I discuss this inequality in $\S$ 7.3.4 of these notes.  Unfortunately the slightly later proof that $f'' \geq 0$ implies this inequality is -- as I've just noticed -- faulty.  Instead of the above inequality I derive the weaker inequality $\frac{f(x)-f(a)}{x-a} \leq \frac{f(b)-f(x)}{b-x}$.  (In fact convex functions satisfy the three-secant inequality $\frac{f(x)-f(a)}{x-a} \leq \frac{ f(b)-f(a)}{b-a} \leq \frac{f(b)-f(x)}{b-x}$.  Please draw a picture!)
So I feel fortunate that Ted Shifrin has sketched an alternate proof.  Let's flesh it out.  Assuming $a = f(a)  = 0$ and $f'' > 0$, we must show that the function $g(x) = \frac{f(x)}{x}$ is strictly increasing on $(0,\infty)$.  Its derivative is
$g'(x) = \frac{xf'(x) - f(x)}{x^2}$,
so we want $h(x) = xf'(x) - f(x) > 0$ for all positive $x$; then $g' > 0$, so $g$ is strictly increasing.  Since $h(0) = 0$, it will be enough to show that $h'(x) > 0$ for all $x > 0$.  And indeed
$h'(x) = f'(x) + xf''(x) - f'(x) = xf''(x) > 0$ for all $x > 0$.
I will fix the proof of Theorem 7.18 in my notes when I get the chance.  (Added: I incorporated into my notes the proof of Ted Shifrin in the comment below, which uses only that $f'$ is increasing.  This is with respect to my private comment.  I am also writing notes on the sequential completion of an ordered field which -- surprise, surprise -- contains some annoying details, so I haven't uploaded the corrected copy yet but I should do so relatively soon.)
A: Consider $g(t)=tf'(t)-f(t)-\frac{1}{2}t^2A $ defined on $[0,x]$ , where $A=2\frac{xf'(x)-f(x)}{x^2}$. Then, $g(0)=g(x)=0. $ By Rolle's theorem, $\exists c\in(0,x)$ such that $g'(c)=0$. It follows that $cf''(c)+f'(c)-f'(c)-cA=0$ and upon simplification, we get:  $f'(x)-\frac{f(x)}{x}=\frac{1}{2}xf''(c)$
A: I know this is an old question, but here is an intuitive solution. $g=x \mapsto f(x)/x$ is the gradient of the line from the origin to $(x,f(x))$. If $g$ is not [strictly] increasing, we have a clockwise triangle $\triangle OPQ$ where $O,P,Q$ are on the curve with $P$ in the middle. But by mean value theorem on both $OP$ and $PQ$ we get $f'$ not being [strictly] increasing. Hence if $f'$ is [strictly] increasing, so is $g$.
