<s>[INST] 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. 
 [/INST] Note that
$$\frac{a^3-b^3}{a^2+ab+b^2}=a-b \implies LHS = \sum_{cyc} \frac{a^3}{a^2+ab+b^2} = \sum_{cyc} \frac{b^3}{a^2+ab+b^2}$$
So we get
$$2LHS = \sum_{cyc} \frac{a^3+b^3}{a^2+ab+b^2}$$
But by rearrangement, $a^3+b^3\ge a^2b+ab^2 \implies 3a^3+3b^3 \ge a^3+2a^2b+2ab^2+b^3 = (a+b)(a^2+ab+b^2)$. 
$$\therefore 2LHS = \sum_{cyc} \frac{a^3+b^3}{a^2+ab+b^2} \ge \frac13\sum_{cyc} (a+b)= \frac23(a+b+c)$$
which is what we wanted to show.
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