a mean value theorem question if $f(x)$ can be differentiated to any degree for $x\in (0,+\infty)$ and $f'(x)>0,f''(x)<0$，if $0<a<b$，acordding to mean value theorem we have $\displaystyle\exists \xi\in(a,b),st.\frac{f(b)-f(a)}{b-a}=f'(\xi)$ ,
prove: $\displaystyle \xi<\frac{a+b}{2}$
And I think that's pretty obvious by looking at the graph of the function,but i don't know how to prove it in math words.

Let $f(x) = \arctan x$, $a={1 \over 8}, b={1 \over 2}$. Since $f''(x) <0$ we see that $f'$ is strictly decreasing, so to provide a counterexample, it is sufficient to show that
${f(b)-f(a) \over b-a} \le f'({a+b \over 2})$ or
${8 \over 3} (\arctan { 1\over 2} - \arctan { 1\over 8}) \le {256 \over 281}$.
Evaluating give $0.904... \le 0.911...$.
