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

real-analysis calculus derivatives

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

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.


We may suppose $a \leq b$. First note that $a=\frac a b b+(1-\frac a b) f(0)$. 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 $\frac a {b-a} f(b) \leq \frac b {b-a} f(a)$

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