<s>[INST] Proving $ \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} \ge \frac{2}{3} (a^2+b^2+c^2)$ for $a, b, c > 0$ 
For $a,b,c>0$, I have to prove that $$ \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} \ge \frac{2}{3} (a^2+b^2+c^2).$$

We have:
$$\begin{align} \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} 
&= \sum_{cyc}^{} \frac {a^4}{a^2+ab+b^2} + \sum_{cyc}^{}\frac {a b^3}{a^2+ab+b^2} \\[4pt]
&\ge \frac{(a^2+b^2+c^2 )^2}{a^2+b^2+c^2+ab+bc+ca} + \sum_{cyc}^{}\frac {a b^3}{a^2+ab+b^2} \\[4pt]
&\ge \frac{1}{2} (a^2+b^2+c^2) + \sum_{cyc}^{}\frac {a b^3}{a^2+ab+b^2}
\end{align}$$
but I can't move on.
 [/INST] Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that:
$$45(u^2-uv+v^2)a^6+18(5u^3-3u^2v+3uv^2+5v^3)a^5+$$
$$+3(23u^4-u^3v-6u^2v^2+59uv^3+23v^4)a^4+$$
$$+3(8u^5+9u^4v-18u^3v^2+34u^2v^3+43uv^4+8v^5)a^3+$$
$$+3(u^6+6u^5v-8u^4v^2+28u^2v^4+12uv^5+v^6)a^2+$$
$$+(3u^5+2u^4v-17u^3v^2+19u^2v^3+20uv^4+3v^5)uva+$$
$$+u^2v^2(u^4-2u^2v-u^2v^2+4uv^3+v^4)\geq0,$$ which is obvious.
</s>