$f(x)>0, f''(x)>0$. Prove: $\int_a^b f(x) dx > (b-a) f(\frac{a+b}{2})$ On $[a,b]$ function $f$ is differentiable for arbitrary order, and $f(x)>0, f''(x)>0$. Prove: $\int_a^b f(x) dx > (b-a) f(\frac{a+b}{2})$.
I first try Taylor expansion at $x_0=\frac{a+b}{2}$, and drop higher order term ($(x-x_0)^3$ terms). But this works only locally at the neighborhood of $x_0$. How to prove this inequality on a finite interval? Thank you.
 A: You have the Taylor expansion around the correct point, you just need to only go to the second order term instead of the third order term. That is use that
$$f(x)  = f\Big(\frac{a+b}{2}\Big) + f'\Big(\frac{a+b}{2}\Big)\Big(x-\frac{a+b}{2}\Big) + \frac{1}{2}f''(\xi)\Big(x- \frac{a+b}{2}\Big)^2$$
Then since we assume that $f'' > 0$, it follows that for all $x\in (a,b)$ that
$$
f(x)  > f\Big(\frac{a+b}{2}\Big) + f'\Big(\frac{a+b}{2}\Big)\Big(x-\frac{a+b}{2}\Big) 
$$
If you now integrate $x$ over $[a,b]$ you should get the result.
A: Let $L(x) = f'\left(\frac{a+b}{2}\right)\left(x - \frac{a+b}{2}\right) + f\left(\frac{a+b}{2}\right)$. Then, $L$ describes the line tangent to $f$ at $x=\frac{a+b}{2}$. Note that:
$$\int_{a}^{b}L(x)\ dx = (b-a)f\left(\frac{a+b}{2}\right)$$
Now, consider $g(x) = f(x)-L(x)$. Note that $g''(x) > 0$ because $L''(x)=0$ and $f''(x)>0$. Additionally, note that $ g\left(\frac{a+b}{2}\right) = g'\left(\frac{a+b}{2}\right) = 0$.
Note that $g'(x)$ is an increasing function because $g''(x)>0$. Then, since $g'\left(\frac{a+b}{2}\right) = 0$, we must have $g'(x) > 0$  for $x>\frac{a+b}{2}$. Repeating this argument for $g(x)$, we can see that $g(x)>0$ for $x>\frac{a+b}{2}$.
Similarly, since $g'(x)$ is increasing and $g'\left(\frac{a+b}{2}\right) = 0$, we must have $g'(x)<0$ for $x<\frac{a+b}{2}$. This shows that $g(x)$ is decreasing for $x<\frac{a+b}{2}$. Combining this with $g\left(\frac{a+b}{2}\right) = 0$, we see that $g(x) > 0$ for $x<\frac{a+b}{2}$.
We can conclude that $g(x) > 0$ for all $x\neq \frac{a+b}{2}$. Thus:
$$\int_{a}^{b}g(x)\ dx = \int_{a}^{b}f(x) - L(x)\ dx > 0$$
$$\int_{a}^{b}f(x)\ dx > \int_{a}^{b}L(x)\ dx$$
$$\boxed{\int_{a}^{b}f(x)\ dx > (b-a)f\left(\frac{a+b}{2}\right)}$$
An intuitive but less rigorous way to understand this is that $f(x)$ is always above $L(x)$ because convex functions (defined by $f''(x)>0$) "curve" upwards from their tangent lines.

A: That is just the Hermite-Hadamard inequality. Up to rescaling we may assume $[a,b]=[-1,1]$ and consider the following: $f(x)$ and $f(-x)$ are strictly convex functions, so $f(x)+f(-x)=g(x)$ is a strictly convex function.
$$ \int_{-1}^{1}f(x)\,dx = \int_{-1}^{1}f(-x)\,dx = \frac{1}{2}\int_{-1}^{1}g(x)\,dx $$
Since $g$ is strictly convex and even, $g(0)$ is necessarily a minimum for $g(x)$ over $[-1,1]$, hence
$$ \int_{-1}^{1} f(x)\,dx > g(0) $$
and by scaling back we get the claim. Actually you do not even need that $f''$ exists, you just need convexity.
