We observe the miraculous identity \[\frac{1}{b^3+4} \ge \frac14 - \frac{b}{12} \]since $12-(3-b)(b^3+4) = b(b+1)(b-2)^2 \ge 0$. Moreover, \[ ab+bc+cd+da = (a+c)(b+d) \le \left( \frac{a+b+c+d}{2} \right)^2 = 4. \]Thus \[ \sum_{\text{cyc}} \frac{a}{b^3+4} \ge \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12} \ge 1 - \frac13 = \frac23. \]This minimum $\frac23$ is achieved at $(a,b,c,d) = (2,2,0,0)$ and permutations