USAMO 2017 Q6

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 $2/3$ is achieved at $(a,b,c,d)$ = $(2,2,0,0)$ and permutations.

$\frac{a}{b^3+4}$ - $\frac{a}{4}$ =- $\frac{a}{4(b^3+4)}$ THEN SHOW BY CONCEPT OF MAXIMA OR MINIMA OR ELSE $b*(b+1)*(b-2)^2>=0$ THAT $\frac{b^3}{b^3+4}$ < $\frac{b}{3}$ THEN USE SIMPLE AM GM AND CONCLUDE ....ANS WILL BE $\frac{2}{3}$ AT (2,2,0,0).