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 $a+b+c+d=4$

Question

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

Answer 1

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!