<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] 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:
$$ \sum {{{{a^3}} \over {{a^2} + ab + {b^2}}}}  = {1 \over 2}\sum {{{{a^3} + {b^3}} \over {{a^2} + ab + {b^2}}}}  \ge {{a + b + c} \over 3}$$
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