Problem: 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 $a+b+c+d=4$.

\end{enumerate}

Solution: The minimum $\frac23$ 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
\[ \frac{1}{b^3+4} \ge \frac14 - \frac{b}{12} \]
since $12-(3-b)(b^3+4) = b(b+1)(b-2)^2 \ge 0$.
Moreover,
\[ ab+bc+cd+da = (a+c)(b+d)
  \le \left( \frac{(a+c)+(b+d)}{2} \right)^2 = 4.  \]
Thus
\[ \sum_{\text{cyc}} \frac{a}{b^3+4}
  \ge \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12}
  \ge 1 - \frac13 = \frac23. \]

\begin{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.
\end{remark*}