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. 
 A: 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.
A: 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.
A: First, we find : 
$$\sum {{{{a^3}} \over {{a^2} + ab + {b^2}}}}  = \sum {{{{a^3} - {b^3} + {b^3}} \over {{a^2} + ab + {b^2}}}}  = \sum {\left( {a - b} \right)}  + \sum {{{{b^3}} \over {{a^2} + ab + {b^2}}}}  = \sum {{{{b^3}} \over {{a^2} + ab + {b^2}}}} $$
So :
$$\sum {{{{a^3}} \over {{a^2} + ab + {b^2}}}}  = {1 \over 2}\sum {{{{a^3} + {b^3}} \over {{a^2} + ab + {b^2}}}} $$
Since :
$${\left( {a - b} \right)^2} \ge 0 \Longrightarrow 2{a^2} + 2{b^2} - 4ab \ge 0 \Longrightarrow 3\left( {{a^2} - ab + {b^2}} \right) \ge {a^2} + ab + {b^2} \\
\Longrightarrow {{{a^2} - ab + {b^2}} \over {{a^2} + ab + {b^2}}} \ge {1 \over 3} \Longrightarrow {1 \over 2} \cdot {{{a^3} + {b^3}} \over {{a^2} + ab + {b^2}}} \ge {{a + b} \over 6} $$
So we get:
$$ \sum {{{{a^3}} \over {{a^2} + ab + {b^2}}}}  = {1 \over 2}\sum {{{{a^3} + {b^3}} \over {{a^2} + ab + {b^2}}}}  \ge {{a + b + c} \over 3}$$
A: We need to prove that
$$\sum_{cyc}\left(\frac{a^3}{a^2+ab+b^2}-\frac{a}{3}\right)\geq0$$ or
$$\sum_{cyc}\frac{a(2a+b)(a-b)}{a^2+ab+b^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{a(2a+b)(a-b)}{a^2+ab+b^2}-(a-b)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(a+b)}{a^2+ab+b^2}\geq0.$$
Done!
