Consider points $a,b,c$ (not on a line) and $x_1,...,x_n$ in $\Bbb{R}^2$. I am looking for a necessary and sufficient condition in terms of the geometric configuration of these points such that for any convex function $f : \Bbb{R}^2 \longrightarrow \Bbb{R}$, $\frac{f(x_1)+\cdots+f(x_n)}{n} \leq \frac{f(a)+f(b)+f(c)}{3}$. (As a guess, I think the points $x_i$ must be in the triangle determined by $a,b,c$). If you know a reference or an idea please let me know.
The necessary and sufficient conditions are: $(1/n)\sum x_j=(1/3)(a+b+c)$, and all $x_j$ lie inside the closed triangle $(a,b,c)$. Proof of sufficiency. For every affine function, we have equality. Let $f$ be convex. Then there exists a unique affine function $g$ which matches $f$ at $a$, $b$ and $c$. And we have $f\leq g$ in the triangle $(a,b,c)$. Thus $$\sum f(x_j)/n\leq\sum g(x_j)/n=(g(a)+g(b)+g(c))/3=(f(a)+f(b)+f(c))/3.$$ Necessity. If some point is outside, it is easy to construct a piecewise affine function which is zero on the triangle and positive outside, so the inequality is violated. If $A=(x_1+...+x_n)/n$, does not coincide with $B=(a+b+c)/3$, then it is easy to see that there exists an affine function which violates the inequality: just take any affine $g$ such that $g(A)>g(B)$. This has a straightforward generalization to any dimension. 

