Prove that $f$ is increasing if and only if a given inequality holds Let $f: [0, \infty) \to \mathbb{R}$ be a continuous function. Prove that $f$ is increasing if and only if:
$$\int_a^b f(x) dx \leq bf(b) - af(a), \, \forall \, \, 0 \leq a \leq b.$$
I have no difficulties in proving that if $f$ is increasing then the inequality holds. But I haven't figured out yet how to prove it the other way around, that is knowing the inequality and proving that $f$ is increasing.
Thank you!
 A: Prove that the integral equation implies that $f$ is increasing via contraposition, viz. prove that if $f$ is not increasing then there exist
$a_0$ and $b_0$ such that $$\int_{a_0}^{b_0} f(x) dx > b_0f(b_0) - a_0f(a_0).$$
Indeed, if $f$ is differentiable, not increasing and not everywhere constant, then there exists an interval $[a_0, b_0]\in\mathbb{R^+}$ such for all $x\in [a_0, b_0]$ we have that $$f(a_0)\geq f(x) \geq f(b_0)\mbox{ and } f(a_0)> f(b_0).$$
Therefore, we have that $$\int_{a_0}^{b_0} f(x) dx \geq f(b_0)(b_0 - a_0)= f(b_0)b_0 - f(b_0)a_0 > f(b_0)b_0 -f(a_0)a_0.$$
A: HINTS:
Show that, WLOG, we can assume $a = 0$. Work with
$$\int_0^{b} f(x)dx \leq bf(b)$$
It should make the result more clear.
A: Note that if $f(a) > f(b) $ then by continuity we can choose $c\in(a, b] $ such that $f(x) > f(c) $ for all $x\in [a, c) $ and hence $$\int_{a} ^{c} f(t) \, dt>(c-a) f(c) $$ And given condition on $f$ implies that integral above is not greater than $cf(c) - af(a) $. It now follows that $f(a) <f(c) $ which is contrary to $f(a) >f(c) $.  Thus we must have $f(a) \leq f(b) $. 
