Proving a convexity inequality

Question

Given $f: \mathbb{R} \to \mathbb{R}$ convex, show that: $$ \frac{2}{3}\left(f\left(\frac{x+y}{2}\right) + f\left(\frac{z+y}{2}\right) + f\left(\frac{x+z}{2}\right)\right) \leq f\left(\frac{x+y+z}{3}\right) + \frac{f(x) + f(y) + f(z)}{3}.$$

I have tried some ideas, such as transforming it into $$ f\left(\frac{x+y}{2}\right) + f\left(\frac{z+y}{2}\right) + f\left(\frac{x+z}{2}\right) - 3f\left(\frac{x+y+z}{3}\right)\\ \leq f(x) + f(y) + f(z) - f\left(\frac{x+y}{2}\right) - f\left(\frac{z+y}{2}\right) - f\left(\frac{x+z}{2}\right) $$ (which graphically seemed intuitive) and using that such an $f$ lies above its tangents, but did not succeed… Ideas are welcome :)

Answer 1

Let $x\geq y\geq z$.

Consider two cases.

1. $x\geq y\geq\frac{x+y+z}{3}\geq z$.

From here we see that $2y\geq x+z$ and easy to check that $$\left(\frac{x+y+z}{3},\frac{x+y+z}{3},\frac{x+y+z}{3},z\right)\succ\left(\frac{x+z}{2},\frac{x+z}{2},\frac{y+z}{2},\frac{y+z}{2}\right),$$ which by Karamata gives $$3f\left(\frac{x+y+z}{3}\right)+f(z)\geq2f\left(\frac{x+z}{2}\right)+2f\left(\frac{y+z}{2}\right).$$ Thus, it's enough to prove that $$f(x)+f(y)\geq2f\left(\frac{x+y}{2}\right),$$ which is Jensen.

2. $x\geq \frac{x+y+z}{3}\geq y\geq z$.

This case is similar. Try to kill it by yourself.