If $f(0)=0$ and $f'$ is increasing, then $\frac{f(x)}{x}$ is increasing. Problem: Suppose $f$ is continuous for $x\ge 0$, differentiable for $x>0$, $f(0)=0$, and $f'$ is monotonically increasing.
Define $g(x)=\frac{f(x)}{x}$ for $x>0$. Prove that $g$ is monotonically increasing.
Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 6. 
 A: Note that $g$ is differentiable everywhere it is defined. Fix $x, y\in\mathbb{R}$ with $y>x$. By the mean-value theorem, for some $t\in(x,y)$, we have
$$\frac{g(y)-g(x)}{y-x} = g'(t)= \frac{tf'(t)-f(t)}{t^2}.$$
Because $y-x$ is positive, it suffices to show that $$h(x)=\frac{xf'(x)-f(x)}{x^2}$$
is nonnegative. This reduces to showing that 
$$f'(x)\ge \frac{f(x)}{x}$$
for $x>0$.
Consider the mean-value theorem applied to $[0,x]$ For some $t\in(0,x)$,
$$\frac{f(x)-f(0)}{x}=f'(t),$$
and the result follows because $f(0)=0$ and $f'$ is monotonically increasing.
A: By assumption, $f$ is a convex function. Then $$g(x)=\frac{f(x)-f(0)}{x-0}$$ must be increasing in $x$, by a standard property of convex functions of one variable.
A: $${f(x)\over x}=\int_0^1 f'(t\, x)\ dt\qquad(x>0)\ .$$
A: The condition that $f'$ is increasing implies that $f$ is convex. Geometrically, the statement to be proved is that for $0 < x < y$ the slope of the chord from $(0,0)$ to $(x,f(x))$ is less than or equal to the slope of the chord from $(0,0)$ to $(y,f(y))$. Thus we
expect to use the definition of convexity for the three points $0,x$, and $y$:
$$f({1 -x  \over y}0 + {x \over y}y) \leq {1 -x  \over y}f(0) + {x \over y}f(y)$$
Since $f(0) = 0$, this is the same as 
$$f(x) \leq {x \over y} f(y)$$
This in turn is the same as
$${f(x) \over x} \leq {f(y) \over y}$$
