Application of Chebyschev inequality 
I want to prove the inequality above.  On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is

We can square the left hand side of our inequality with the term to its right to get
$$\frac{a+b+c}{3}\cdot \frac{a+b+c}{3}\geq \frac{ca+b^2+ca}{3}$$
My problem
I would've wanted to get $\dfrac{ab+bc+ca}{3}$ on the right instead.  Did I apply the inequality wrong, or does it follow that $\dfrac{ab+bc+ca}{3}$ is less than or equal to $\dfrac{a+b+c}{3}\cdot \dfrac{a+b+c}{3}$?

As you've noted here the use of the Chebyshev's Inequality doesn't help much. However we can make use of the AM-GM inequality.
Squaring both sides and expanding them we get that the inequality is equivalent to: $(a+b+c)^2 \ge 3(ab+bc+ca)$, which is equivalent to $a^2 + b^2 + c^2 \ge ab + bc + ca$.
The last equality is true, as we have that $a^2 + b^2 \ge 2ab$, $b^2 + c^2 \ge 2bc$ and $c^2 + a^2 \ge 2ac$. Add the inequalities to get the final answer.
