How prove this? $\dfrac{\int_{a}^{b}f(x)dx}{b-a} \ge \dfrac{\max f(x)}{2} , aLet $f(x)\in R$ be a concave function then show that
$$\dfrac{\int_{a}^{b}f(x)dx}{b-a} \ge \dfrac{\max f(x)}{2} , x\in [a,b],a<b$$
I have
$$M=max f(x)$$
$$\int_{a}^{b}f(x)dx \le \int_{a}^{b}M$$
$$\dfrac{\int_{a}^{b}f(x)dx}{b-a}\le M$$
 A: Assuming that $f \ge 0$, we can proceed as follows. 
Let $c$ be the point where $f$ attains the maximum. It's clear by staring at the picture that area of $f$ under graph $\int_a^b f(x)dx$ is greater than the area of the triangle with vertices $(a,0)$, $(c,f(c))$, $(b,0)$, which is $\frac{(b-a)\max f(x)}{2}$. This can be rigorously proved by integrating 
$$f(tx+(1-t)y) \ge tf(x) + (1-t)f(y) \text{ for $t$ over $[x,y]$}$$
 where $(x,y) = (a,c)$ and $(c,b)$.
A: I am not sure if there is any easy way to do this, but there is an inequality due to Favard (J Favard, Sur les valeurs moyennes, Bull. Sci. Math., 57 (1933), pp. 54–64 (2)) :

If $f:[a,b]\to \mathbb{R}_{+}$ is a continuous concave function taking non-negative values, and $p>1$, then
  $$
\left ( \frac{1}{b-a} \int_a^b f^p(x)dx \right )^{1/p} \leq \frac{2}{(p+1)^{1/p}}\left ( \frac{1}{b-a}\int_a^b f(x)dx \right )
$$

Now if you let $p\to \infty$, then you will get what you want. (Note: Here we require that $f$ be non-negative though)
A: I don't think the claim is correct. take $f(x)=-x^2$ in $[-9,1]$ and you'll get $$\int_{-9}^1-x^2=\frac {x^3} 3\mid_{-9}^{1}=-243.33\overset{?}{\ge}4\cdot 0=0$$ which is not correct.
EDIT: If you add $f\in \mathbb R_+$ than look at the answer of ferharld which will be correct in this case.
A: This might look like a fundamental theorem of calculus question, but it isn't.  FTC requires the function be smooth, and $f$ isn't given to be smooth here.  Furthermore, it actually isn't necessary for $f$ to be smooth for the statement to be true.
Somehow you are going to have to use the definition of concavity to establish that $\int_a^b f(x) - L(x) dx \ge 0$ for a line $L$ passing through $(a, f(a))$ and $(b, max f)$.  Combine that result with $f(a) \ge 0$ and you get the final result you want.
A: $f$ has to be positive. The simplest way to see the inequality is with a graph.

(source: subefotos.com)
