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.

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
$f\left(\displaystyle\frac{x_1+x_2+x_3}3\right) \le \displaystyle\frac{f\left(\frac{x_1+x_2}2\right) + f\left(\frac{x_2+x_3}2\right) + f\left(\frac{x_3+x_1}2\right)}{3} \le\displaystyle\frac{f(x_1)+f(x_2)+f(x_3)}{3}$
