Why is $f'(x) < f(x)/x$ for $f''(x)<0$ How can it be shown formally that for all $x>0$


*

*$f(x) > 0$,

*$f'(x) > 0$, and

*$f''(x) < 0$


imply
$f'(x) < \frac{f(x)}{x}$?
My (somewhat sloppy) intuition is this: since $f'(x)$ is decreasing, any marginal increase $f'(x_0)$, must be smaller than any previous marginal increase for $x<x_0$ and thus, smaller than the average of all previous marginal increases, which is given by $\frac{f(x_0)}{x_0}$.
 A: Let $ F(x) = xf'(x)-f(x)$. Then $F'(x) = f'(x) + xf''(x) - f'(x) = x f''(x) \lt 0.$  Thus $F(x)$ is decreasing for $x \gt 0.$   
Can you conclude?
A: If the conditions given only hold at a specific $x$, then the statement is not true.
Take $f(x) = 3-(x-1)^2$. Consider $x=1-\epsilon$, with $\epsilon>0$. Then by choosing $\epsilon$ small enough, we can find a point $1-\epsilon$ such that $f'(1-\epsilon) = 2 \epsilon < { f(1-\epsilon) \over 1 - \epsilon } = { 3 - \epsilon^2 \over 1 - \epsilon }$.
If the conditions hold for all $x >0$ (and $f$ is twice differentiable for $x>0$), then we note that since $f(x) >0$ and $f'(x) >0$, we have $\lim_{x \downarrow 0} f(x)$ exists and is non-negative.
So we may take $f$ to be defined at $x=0$, with value $f(0)= \lim_{x \downarrow 0} f(x)$. The resulting $f$ is continuous for $x \ge0$ and twice differentiable for $x>0$.
Since $f''(x) <0$, we see that $f'(x)$ is strictly decreasing.
By the mean value theorem, for $x>0$ we have $f(x)-f(0) = f'(\xi)x$ for some $\xi \in (0,x)$. Hence
$f'(x) < f'(\xi) = { f(x)-f(0) \over x } \le { f(x) \over x }$ (since $f(0) \ge 0$).
