# How prove this? $\dfrac{\int_{a}^{b}f(x)dx}{b-a} \ge \dfrac{\max f(x)}{2} , a<b$

Let $f(x)\in R$ 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

$$M=max f(x)$$

$$\int_{a}^{b}f(x)dx \le \int_{a}^{b}M$$

$$\dfrac{\int_{a}^{b}f(x)dx}{b-a}\le M$$

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Are you possibly leaving out the condition that f(x) is positive? – DanielV Oct 5 '13 at 5:31

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 $(c,b)$.

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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)

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The requirement that $f\geq 0$ is good, because otherwise it can be false. E.g. $[a,b]=[0,1]$, $f(x)=x-1$. – Jonas Meyer Oct 5 '13 at 5:32

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.

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He edited the question and added that is was concave. – DanielV Oct 5 '13 at 5:30
@DanielV: I don't think that is true. It said concave from the beginning. Based on the edit history you can see that it was added at latest within 5 minutes of posting. If it was added 5 minutes after posting, that is still 26 minutes before this answer came. – Jonas Meyer Oct 5 '13 at 5:33
I edited my answer. – user65985 Oct 5 '13 at 5:35
The updated example shows why the assumption that $f\geq 0$ should be added to the question. – Jonas Meyer Oct 5 '13 at 5:35
but he haven't added it yet and it completely changes the question. – user65985 Oct 5 '13 at 5:36

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.

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$f$ has to be positive. The simplest way to see the inequality is with a graph.

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