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.

This has a straightforward generalization to any dimension.

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.

The necessary and sufficient conditions are: $\sum x_j=a+b+c$, 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 $A$, the baricenter of $x_j$, does not coincide with $B$, the baricenter of $a,b,c$, then it is easy to see that there exists an affine function which violates the inequality.

This has a straightforward generalization to any dimension.

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.

This has a straightforward generalization to any dimension.