Beautiful cyclic inequality Prove that cyclic sum of $\displaystyle \sum_{\text{cyclic}} \dfrac{a^3}{a^2+ab+b^2} \geq \dfrac{a+b+c}{3}$ , if $a, b, c > 0$
I'm really stuck on this one.  Tried some stuff involving QM> AM(because the are positive) but can't derive the needed  ,can't proceed from it. 

Here is another way, using Cauchy-Schwarz inequality:
$$\sum_{cyc} \frac{a^4}{a(a^2+ab+b^2)} \ge \frac{(a^2+b^2+c^2)^2}{\sum_{cyc} a(a^2+ab+b^2)} = \frac{(a^2+b^2+c^2)^2}{(a+b+c)(a^2+b^2+c^2)}$$
So it is enough to show that
$$a^2+b^2+c^2 \ge \frac{(a+b+c)^2}3$$
which is easy again using Cauchy-Schwarz.
