convex function inequality similr to the Hermite-Hadamard inequality

Question

Given a function $f: [a,b] \to \mathbb{R}$, that is convex and continuous, prove that the following inequality:

$$\frac{2}{b-a}\int_a^b f(x) \,dx\le\frac{f(a)+f(b)}{2}+f\left(\frac{a+b}{2}\right).$$

I tried using the Hermite-Hadamard inequality because I thought it had a similar form, but nothing came out.

Answer 1

The picture says it all. Use the formula for the area of a trapezoid.

UPD: I have no idea why my hint got downvoted, so here is the full solution.

As we can see in the picture due to function convexity and continuity we have the following area inequality: $$\int_{a}^b f(t)dt \le S_{ABED} + S_{BCFE}$$ Both $ABED$ and $BCFE$ are trapeziods, so $$S_{ABED} = \frac{f(a)+f\left(\frac{a+b}{2}\right)}{2}\cdot\frac{b-a}{2}$$ $$S_{BCFE} = \frac{f\left(\frac{a+b}{2}\right) + f(b)}{2}\cdot\frac{b-a}{2}$$ Now we have: $$\int_a^b f(t)dt \le \frac{b-a}{2}\left(f\left(\frac{a+b}2\right) + \frac{f(a)+f(b)}{2}\right)$$ QED

Answer 2

Step 1

For $a \leq s < u < t \leq b$, the following holds.

$$f(u)+f(s+t-u) \leq f(s)+f(t)$$

I’ll leave this as an exercise. Just use the definition of convexity.

Step 2

For $a \leq s < t \leq b$, the following holds.

$$\frac{1}{t-s}\int_{s}^{t}f(x) \mathrm{d}x \leq \frac{f(s)+f(t)}{2}$$

This is because of the following.

$$2\int_{s}^{t}f(x) \mathrm{d}x = \int_{s}^{t} \{ f(x)+f(s+t-x) \} \mathrm{d}x$$ $$\leq \int_{s}^{t} \{ f(s)+f(t) \} \mathrm{d}x = (t-s)\{ f(s)+f(t) \}$$

Step 3

Add the following.

$$\frac{2}{b-a}\int_{a}^{\frac{a+b}{2}}f(x) \mathrm{d}x \leq \frac{f(a)+f(\frac{a+b}{2})}{2}$$

$$\frac{2}{b-a}\int_{\frac{a+b}{2}}^{b}f(x) \mathrm{d}x \leq \frac{f(\frac{a+b}{2})+f(b)}{2}$$