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.
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.
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.
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.
