Given $a+b+c+d=4$, find the minimum value of $\Sigma_{cyc}\frac{a}{b^3+4}$

Question

Given $a+b+c+d=4$, find the minimum value of $\Sigma_{cyc}\frac{a}{b^3+4}$ I'm pretty sure this has something to do with Holder's inequality, but I don't know how to solve this. By guessing I found $a=2,b=2,c=d=0$ the smallest solution with $\Sigma_{cyc}\frac{a}{b^3+4}=\frac{2}{3}$

Answer 1

Does not exist. Try $a>0$,$c>0$, $d>0$ and $b\rightarrow-\sqrt[3]4^-.$

For positive variables by AM-GM we obtain: $$\sum_{cyc}\frac{a}{b^3+4}=1+\sum_{cyc}\left(\frac{a}{b^3+4}-\frac{a}{4}\right)=1+\sum_{cyc}\frac{-ab^3}{4(b^3+4)}=$$ $$=1+\sum_{cyc}\frac{-ab^3}{2(b^3+b^3+8)}\geq1+\sum_{cyc}\frac{-ab^3}{2\cdot3\sqrt[3]{b^6\cdot8}}=1-\frac{1}{12}\sum_{cyc}ab=$$ $$=1-\frac{(a+c)(b+d)}{12}\geq1-\frac{\left(\frac{a+c+b+d}{2}\right)^2}{12}=\frac{2}{3}.$$