Suppose h is twice differentiable on $[0,2b]$ for some $b > 0$ with $h(0) = 0$ and $h''(x) < 0$.

Question

The question asks us to show that $h(x+y) \le h(x) + h(y)$ for all $x,y \in [0,b]$, I move $h(x)$ to the left side and divide $y$ on the both side. Like $$\frac{h(x+y)-h(x)}{y} \le \frac{h(y)}{y}$$, and I try to take limits on the both side, but it seems not working. Can anyone give me some tips? Thank you very much!

Answer 1

Assume that $y<x$. $h''(x) <0$ implies that $h'$ is decreasing. Note that by MVT $\frac {h(x+y)-h(x)} {y}=f'(t)$ for some $t \in (x,x+y)$ and $\frac {h(y)} y=\frac {h(y)-h(0)} y=h'(s)$ for some $s \in (0,y)$. Since $s <y <x<t$ we see that $h'(s) \geq h'(t)$ which gives the desired inequlaity. When $x <y%=$ just switch the roles of $x$ and $y$.