Find the minimum possible value of
$$
\frac{a}{b^{3}+4}+\frac{b}{c^{3}+4}+\frac{c}{d^{3}+4}+\frac{d}{a^{3}+4}
$$
given that $a, b, c, d$ are nonnegative real numbers such that $a+b+c+d=4$.
The minimum $\frac{2}{3}$ is achieved at $(a, b, c, d)=(2,2,0,0)$ and cyclic permutations.
The problem is an application of the tangent line trick: we observe the miraculous identity
$$
\frac{1}{b^{3}+4} \geq \frac{1}{4}-\frac{b}{12}
$$
since $12-(3-b)\left(b^{3}+4\right)=b(b+1)(b-2)^{2} \geq 0$. Moreover,
$$
a b+b c+c d+d a=(a+c)(b+d) \leq\left(\frac{(a+c)+(b+d)}{2}\right)^{2}=4 .
$$

Thus
$$
\sum_{\text {cyc }} \frac{a}{b^{3}+4} \geq \frac{a+b+c+d}{4}-\frac{a b+b c+c d+d a}{12} \geq 1-\frac{1}{3}=\frac{2}{3} .
$$

Remark. The main interesting bit is the equality at $(a, b, c, d)=(2,2,0,0)$. This is the main motivation for trying tangent line trick, since a lower bound of the form $\sum a(1-\lambda b)$ preserves the unusual equality case above. Thus one takes the tangent at $b=2$ which miraculously passes through the point $(0,1 / 4)$ as well.