Problem using Jensen's Inequality For some convex function $f$, and elements $x_{1},x_{2},x_{3}$ of its domain, show that:
$$f(x_{1})+f(x_{2})+f(x_{3})+f\left( \frac{1}{3}(x_{1}+x_{2}+x_{3}\right)\ge \frac{4}{3}\left( f\left(\frac{x_{1}+x_{2}}{2} \right)+f\left(\frac{x_{2}+x_{3}}{2} \right)+f\left(\frac{x_{3}+x_{1}}{2} \right) \right)$$
So far I have that the left hand side is greater than or equal to:
$$4f\left(\frac{x_{1}+x_{2}+x_{3}}{3}\right)=4f\left(\frac{x_{1}+x_{2}}{6}+\frac{x_{2}+x_{3}}{6}+\frac{x_{1}+x_{3}}{6}\right)$$
by Jensen, which we can say is less than or equal to the right hand side (again, by Jensen). My problem is that this doesn't imply the conclusion above, since $a\ge b\le c$ doesn't tell us anything about the relationship between $a$ and $c$. This makes me think I've made a mistake, which I would be very grateful if someone could spot.
 A: The following function $f \colon \mathbb R^2 \to \mathbb R$ is a counterexample.
Consider the equilateral triangle $T$ given by the points $(\cos\varphi_i,\sin\varphi_i)$ for $\varphi_1=0$, $\varphi_2=\frac{2\pi}3$, $\varphi_3=\frac{4\pi}3$. (To be more precise, we define the set $T$ only as the boundary of this triangle, and not the interior.)
We put
$f(a,b)=\alpha$ for the points belonging to $\alpha T$. (The triangle, which is $\alpha$-times larger.)
This function is convex. (It is the Minkowski functional of the triangle.)
But if we take $x_i=(\cos\varphi_i,\sin\varphi_i)$ for $i=1,2,3$ we have $f(x_i)=f\left(\frac{x_i+x_j}2\right)=1$ and $f\left(\frac{x_1+x_2+x_3}3\right)=0$. So the above inequality now becomes $3\ge \frac43\cdot 3=4$, which is false. 

The case of functions $f \colon \mathbb R\to\mathbb R$ (which probably was the original intention of this question) is solved as 
Problem 431 in the book Putnam and Beyond by Razvan Gelca, Titu Andreescu p.526.
A: Use these:
$f\left(\displaystyle\frac{x_i+x_j}2\right) \le \displaystyle\frac{f(x_i)+f(x_j)}{2}$, and
$f\left(\displaystyle\frac{y_1+y_2+y_3}3\right) \le \displaystyle\frac{f(y_1)+f(y_2)+f(y_3)}{3}$.
This applied to $y_1=\displaystyle\frac{x_2+x_3}2$,.. we also have
$f\left(\displaystyle\frac{x_1+x_2+x_3}3\right) \le \displaystyle\frac{f\left(\frac{x_1+x_2}2\right) + f\left(\frac{x_2+x_3}2\right) + f\left(\frac{x_3+x_1}2\right)}{3} \le\displaystyle\frac{f(x_1)+f(x_2)+f(x_3)}{3}$
