On account of AM-GM inequality, we have $\sqrt[3]{bcd} \leq \frac{b + c + d}{3}$ and
$$\begin{aligned} \frac{a - \sqrt[3]{bcd}}{a + 3(b + c + d)} &\ge \frac{1}{3} \left( \frac{3a - (b + c + d)}{a + 3(b + c + d)} \right) = \frac{1}{9} \left( \frac{10a - (a + 3(b + c + d))}{a + 3(b + c + d)} \right) \\ &\ge \frac{1}{9} \left( \frac{10a}{a + 3(b + c + d)} - 1 \right) \end{aligned}$$
Therefore,
$$\sum_{cyc} \frac{a - \sqrt[3]{bcd}}{a + 3(b + c + d)} \ge 0 \Leftrightarrow \sum_{cyc} \left( \frac{10a}{a + 3(b + c + d)} \right) \ge 4,$$
or equivalently,
$$\sum_{cyclic} \frac{a}{a + 3(b + c + d)} \ge \frac{2}{5}$$
Let us denote by $S = a + b + c + d$. Then, we have
$$\sum_{cyc} \frac{a}{a + 3(b + c + d)} = \sum_{cyc} \frac{a}{3S - 2a} = \sum_{cyc} \frac{a^2}{3aS - 2a^2} \ge \frac{\left(\sum_{cyc} a\right)^2}{3S^2 - 2 \sum_{cyc} a^2}$$
on account of Bergströn's inequality:
$$\sum_{cyc} \frac{a^2}{x} \ge \left( \sum_{cyc} a \right)^2 / \sum_{cyc} x$$
To prove
$$\frac{\left(\sum_{cyc} a\right)^2}{3S^2 - 2 \sum_{cyc} a^2} \ge \frac{2}{5}$$
we apply Bergströn's inequality again and we get
$$\sum_{cyc} a^2 \ge \frac{1}{4} \left( \sum_{cyc} a \right)^2$$
Then, we have
$$5 \left( \sum_{cyc} a \right)^2 \ge 6S^2 - 4 \sum_{cyc} a^2$$
from which follows $3(a + b + c + d)^2 \ge 0$. This completes the proof.