We have
$$\begin{aligned} \sum_{cyc} \frac{a^4}{a^3 + a^2 b + ab^2 + b^3} - \sum_{cyc} \frac{b^4}{a^3 + a^2 b + ab^2 + b^3} &= \sum_{cyc} \frac{a^4 - b^4}{a^3 + a^2 b + ab^2 + b^3} \\ &= \sum_{cyc} (a - b) = 0 \end{aligned}$$
$$\sum_{cyc} \frac{a^4 + b^4}{a^3 + a^2 b + ab^2 + b^3} \ge \frac{a + b + c + d}{2}.$$
This, however, follows from
$$\frac{a^4 + b^4}{a^3 + a^2b + ab^2 + b^3} \ge \frac{a+b}{4} \Leftrightarrow 4(a^4 + b^4) \ge (a+b)(a^3 + a^2b + ab^2 + b^3).$$
$$\frac{a^4}{a^3 + a^2b + ab^2 + b^3} \ge \frac{5}{8}a - \frac{3}{8}b \Leftrightarrow 3(a^4 + b^4) \ge 2(a^3b + a^2b^2 + ab^3),$$
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}.