55th IMO Team Selection Test

We have $$\begin{array}{rl}\sum _{cyc}\frac{a^4}{a^3+a^{2}b+a{b}^2+b^3}-\sum _{cyc}\frac{b^4}{a^3+a^{2}b+a{b}^2+b^3}& =\sum _{cyc}\frac{a^4-b^4}{a^3+a^{2}b+a{b}^2+b^3}\\ & =\sum _{cyc}(a-b)=0\end{array}$$ $$\sum _{cyc}\frac{a^4+b^4}{a^3+a^{2}b+a{b}^2+b^3}\ge \frac{a+b+c+d}{2}.$$ This, however, follows from $$\frac{a^4+b^4}{a^3+a^{2}b+a{b}^2+b^3}\ge \frac{a+b}{4}\iff 4(a^4+b^4)\ge (a+b)(a^3+a^{2}b+a{b}^2+b^3).$$ $$\frac{a^4}{a^3+a^{2}b+a{b}^2+b^3}\ge \frac{5}{8}a-\frac{3}{8}b\iff 3(a^4+b^4)\ge 2(a^{3}b+a^2{b}^2+a{b}^3),$$ which is obviously true. Therefore, $$ \sum_{cyc} \frac{a^4}{a^3 + a^2b + ab^2 + b^3} \ge \sum_{cyc} \frac{5}{8}a - \frac{3}{8}b = \frac{a+b+c+d}{4}.