Let $f:(0,+\infty)\to\mathbb{R}$ be a continuous concave function such that $$f(x+h)+f(x-h)-2f(x)\leq0 \;\;; h>0.$$
My question is proving the following inequalities $$\frac{f(x+h_2)-f(x)}{{h_2}}-\frac{f(x)-f(x-h_1)}{{h_1}}\leq0\;\;; h_1,h_2>0$$
I wanted to prove with the following way $$f(x+h_1)+f(x-h_1)-2f(x)\leq0 \;\;; h>0.$$ $$f(x+h_2)+f(x-h_2)-2f(x)\leq0 \;\;; h>0.$$ Firstly, subtracting the two above relation after that adding f(x), but I could not get the result. Even I tried to prove this by the following $$f^\prime(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ But we do not that $f$ is differentiable or not. anyone can help me.
Assuming continuity of $f$ we can show that $f(tx+(1-t)y) \geq tf(x)+(1-t)f(y)$ for all $x,y$ and for all $t \in [0,1]$. This, by the way, is the usual definition of a concave function!. For proving this from your definition use induction to prove this when $t$ is of the form $\frac i {2^{N}}$ and then take limits. [Ant real number $t$ is a limit of a sequence of numbers of the form $\frac i {2^{N}}$]. Once this is done observe that $x=\frac {h_1} {h_1+h_2} (x+h_2) +\frac {h_2}{h_1+h_2} (x-h_1)$ and apply above inequality with $t =\frac {h_1} {h_1+h_2}$, $x$ changed to $x+h_2$ and $y$ changed to $x-h_1$. If you simplify this last inequality you will get the answer.
A counterexamle for the original version of the question: It is well know that there are discontinuous functions $f$ from $\mathbb R$ to $\mathbb R$ such that $f(x+y)=f(x)+f(y)$. Such a function satisfies your hypothesis but not the conclusion. So your claim is false in general.