If $f$ is a concave function and $f(0) \geq 0$ then $f(a+b) \leq f(a) + f(b)$ for all $a,b >0$

Question

Well If I've got a nice understanding, this proof involves the use of Rolle's and/or Mean Value Theorems. But I don't even an idea of how to start. We already know that $$f\text{ concave}\Rightarrow f''<0$$ but, then...

Answer 1

$$ \frac{b}{a+b}f(0) + \frac{a}{a+b}f(a+b) \leq f(a) $$

since $f(0) \ge 0$ we have

$$ \frac{a}{a+b}f(a+b) \leq f(a) $$ similarly by interchanging the role $a$ with $b$ we have

$$ \frac{b}{a+b}f(a+b) \leq f(b) $$

Now by adding last two inequality we arrive to the result.

Answer 2

We may suppose $a \leq b$. First note that $a=\frac a b b+(1-\frac a b) 0$. Hence $f(a) \geq \frac a b f(b)$. Next write b as $\alpha a+(1-\alpha)(a+b)$ and apply the definition of concavity. We get $f(a+b) \leq \frac b {b-a} f(b) - \frac a {b-a} f(a) \leq f(a)+f(b)$ because $\frac a {b-a} f(b) \leq \frac b {b-a} f(a)$