Q:

Concave function first-order conditions

Let $g(x)$ be a function defined over the unit interval $[0, 1]$. Suppose that $g (0) = g (1) = 0$, and $g''(x) \le 0$, for any $x \in [0, 1]$.
Show that for any $x \in (0, 1)$, $g'(x) \le \frac{g(x)}{x}$
What I've tried is the mean value theorem, but can't get the right answer.
$g'(c) = \frac{g(x) - g(a)}{x - a}$
$(x-a)g'(c) = g(x) - g(a) \le g(x)$
$g'(c) \le \frac{g(x)}{x - a}$ and $\frac{g(x)}{x - a} \ge \frac{g(x)}{x} $ as $ 1> x > a > 0$ so dividing by $(x - a)$ yields a larger number.
But obviously the inequalities don't quite work.

A:

You are on the right track. Use the fact that $g(0)=0$ and that $g'$ is decreasing, and consider $\dfrac{g(x)-g(0)}{x-0}$, to get the answer

