A symmetrisation trick: Σ a³/(a² + ab + b²)

Idea. Let $S=\sum\frac{a^3}{a^2+ab+b^2}$ and $T=\sum\frac{b^3}{a^2+ab+b^2}$, both over the cyclic pairs $(a,b),(b,c),(c,a)$.

The difference vanishes. Since $a^3-b^3=(a-b)(a^2+ab+b^2)$,
$$S-T=\sum(a-b)=(a-b)+(b-c)+(c-a)=0.$$
So $S=T=\dfrac{S+T}2$ and
$$S=\frac12\sum_{\rm cyc}\frac{a^3+b^3}{a^2+ab+b^2}.$$

A pair estimate. For $a,b>0$,
$$\frac{a^3+b^3}{a^2+ab+b^2}\ge\frac{a+b}3\iff3(a+b)(a^2-ab+b^2)\ge(a+b)(a^2+ab+b^2)\iff2(a-b)^2\ge0.$$

Conclusion. Summing the three pair estimates, $S\ge\frac12\cdot\frac{2(a+b+c)}3=\frac{a+b+c}3$, with equality iff $a=b=c$.

Lesson. When a cyclic sum has numerators and denominators that differ by a nice factor, add the "mirror" sum: the difference telescopes and the symmetrised sum is easier to bound.