Krdytkyu

Suppose $f$ is a continuous function. Prove $f$ is convex iff this inequality holds:
$$
frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})]
$$

if we move all the terms to the left hand side, we'll have:
$$
frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge 0
$$
I tried to prove this statement with this assumption that f is convex. First, I used the definition of convexity for each terms of the right hand side:
$$f(frac{x+y}{2})le frac{1}{2}f(x)+frac{1}{2}f(y)$$
$$f(frac{y+z}{2})le frac{1}{2}f(y)+frac{1}{2}f(z)$$
$$f(frac{z+x}{2})le frac{1}{2}f(z)+frac{1}{2}f(x)$$
$$Longrightarrow f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2}) le f(x)+f(y)+f(z)
$$
Then after using definition for $f(x)+f(y)+f(z)/3$ and replacing above statement in the main inequality, I obtained:
$$
frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3})-frac{2}{3}[f(frac{x+y}{2}) +f(frac{y+z}{2}) +f(frac{z+x}{2})] ge f(frac{x+y+z}{3})-frac{f(x)+f(y)+f(z)}{3}
$$
According to the convexity of $f$, the later expression must be nonpositive but our purpose is to prove it is nonnegative. It seems we encounter with a contradiction! I tried other types of convex combinations, but I just obtained either of parts are greater or smaller than a same term, for example this one:

first part:$$frac{f(x)+f(y)+f(z)}{3} + f(frac{x+y+z}{3}) ge 2f(frac{x+y+z}{3})$$
second part:$$2[frac{1}{3}f(frac{x+y}{2}) +frac{1}{3}f(frac{y+z}{2}) +frac{1}{3}f(frac{z+x}{2})] ge 2f(frac{x+y+z}{3})$$

I think it needs to obtain a tighter bound or maybe some new information about continuous convex functions.

This is not true for functions $f:mathbb R^n to mathbb R$ for $nge2$:

Take a convex function that is zero on the boundary of the triangle $(x,y,x)$ but negative in barycenter $(x+y+z)/3$.