Minimizing $\frac{d}{a^3+4}+\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}$ for nonnegative parameters with $a+b+c+d=4$ 
Minimize
$$\frac{d}{a^3+4}+\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}$$
where $a$, $b$, $c$, $d$ are nonnegative and $a+b+c+d = 4$.

I know the minimum is $2/3$ when, say, $a=b=2$.  I found this using calculus, which is very computational.
Does anybody have an algebraic solution.  I am unable to find one.  Thank you!
Source: https://brilliant.org/problems/volcanic-inequality/?ref_id=1413038

For $(a,b,c,d)=(2,2,0,0)$ we get a value $\frac{2}{3}$.
We'll prove that it's a minimal value.
Indeed, by AM-GM
$$\sum_{cyc}\frac{a}{b^3+4}=\frac{1}{4}\sum_{cyc}\left(a-\frac{ab^3}{b^3+4}\right)=\frac{1}{4}\sum_{cyc}\left(a-\frac{2ab^3}{2b^3+8}\right)\geq$$
$$\geq\frac{1}{4}\sum_{cyc}\left(a-\frac{2ab^3}{3\sqrt[3]{b^6\cdot8}}\right)=\frac{1}{12}\sum_{cyc}(3a-ab).$$
Thus, it remains to prove that
$$\sum_{cyc}(3a-ab)\geq8$$ or
$$a+b+c+d\geq ab+bc+cd+da$$ or
$$(a+b+c+d)^2\geq4(a+c)(b+d),$$
which is AM-GM again.
Done!
