JustPostTaiwanTST wrote: Prove that for any positive integers $ a,b,c,d $ with $ a+b+c+d = 4 $, we have $$ \sum\limits_{cyc}{\frac{a^3}{a^2+ab+b^2}}+\sum\limits_{cyc}{\frac{2ab}{a+b}} \ge 8 $$ Positive integer? Anyway putting $a,b,c,d = 1$, we have $LHS = \frac{4}{3} + 4$, which is less than $8$?

we have $a^3-b^3=(a-b)(a^2+ab+b^2)$,so it's easy to find that $\sum_{cyc}\frac{3a^3}{a^2+ab+b^2}=\sum_{cyc}\frac{3b^3}{a^2+ab+b^2}$. now we proof :$$ \sum\limits_{cyc}{\frac{3a^3}{a^2+ab+b^2}}+\sum_{cyc}\frac{3b^3}{a^2+ab+b^2}+\sum\limits_{cyc}{\frac{4ab}{a+b}} \ge 16 $$actually we have $\frac{3a^3}{a^2+ab+b^2}+\frac{3b^3}{a^2+ab+b^2}\ge \frac{2a^2+2b^2}{a+b}$ so $L.H.S\ge \sum_{cyc}\frac{2a^2+2b^2}{a+b}+\sum\limits_{cyc}{\frac{4ab}{a+b}}=\sum_{cyc}(2a+2b)=16$. so we are done.

(proof for $\frac{3a^3}{a^2+ab+b^2}+\frac{3b^3}{a^2+ab+b^2}\ge \frac{2a^2+2b^2}{a+b}$) is equivalent to $a^4+a^3b+ab^3+b^4\ge 4a^2b^2$ AM-GM.