Let $f$ be a continuous real valued function defined on $[a,b]$. Prove that, if $f$ is convex, then:
$$f\left(\frac{a+b}{2}\right)(b-a) \geq \int_a^b f$$
I am looking for a proof of this that doesn’t involve Riemann sums, but I can’t find any.
Thanks in advance.
On
The inequality is reversed. It is a part of a celebrated Hermite-Hadamard inequality. One of the possible proofs is to consider affine support $\varphi$ at the midpoint $\frac{a+b}{2}.$ Its equation is not needed. Nonetheless, it is $$\varphi(x)=\alpha\left(x-\frac{a+b}{2}\right)+f\left(\frac{a+b}{2}\right)$$ with any $\alpha$ between $f'_-\left(\frac{a+b}{2}\right)$ and $f'_+\left(\frac{a+b}{2}\right)$ which do exist by convexity. It is enough to know that $\varphi(x)\le f(x)$ on $[a,b]$ with the equation at the midpoint. Now integrate $\varphi$ and $f$. The proof is done.
\begin{align*} f\left(\int_{a}^{b}x\cdot\dfrac{dx}{b-a}\right)\leq\int_{a}^{b}f(x)\cdot\dfrac{dx}{b-a}, \end{align*} where $\displaystyle\int_{a}^{b}xdx=\dfrac{1}{2}(b+a)(b-a)$, here I have used Jensen's inequality.