Notice that
$$\sum_{\text{cyc}} \frac{a^3 - b^3}{a^2 + ab + b^2} = \sum_{\text{cyc}} (a-b) = 0.$$
Hence
$$\text{LHS} = \sum_{\text{cyc}} \frac{3a^3 + 3b^3}{2(a^2 + ab + b^2)} + \sum_{\text{cyc}} \frac{2ab}{a+b}.$$
Now
$$\frac{2ab}{a+b} = a+b-\frac{a^2+b^2}{a+b},$$
which means we only need to prove that
$$\sum_{\text{cyc}} \frac{3a^3 + 3b^3}{2(a^2 + ab + b^2)} \geq \sum_{\text{cyc}} \frac{a^2 + b^2}{a+b}.$$
This is true since
$$3(a^3 + b^3)(a+b) \geq 3(a^2 + b^2)^2 \geq 2(a^2 + ab + b^2)(a^2 + b^2).$$