Given $a+b+c+d=4$ To find Minimum value of $\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$

Question

Given $a+b+c+d=4$ where $a,b,c,d \in \mathbb{R^{+}}$

find Minimum value of $$S=\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$$

I have no clue to start...any hint?

Answer 1

By AM-GM $$\sum_{cyc}\frac{a}{b^3+4}=\frac{a+b+c+d}{4}-\sum_{cyc}\left(\frac{a}{4}-\frac{a}{b^3+4}\right)=$$ $$=1-\sum_{cyc}\frac{b^3a}{4\left(\frac{b^3}{2}+\frac{b^3}{2}+4\right)}\geq1-\frac{b^3a}{4\cdot3\sqrt[3]{\left(\frac{b^3}{2}\right)^2\cdot4}}=$$ $$=1-\sum_{cyc}\frac{b^3a}{12b^2}=1-\frac{1}{12}(ab+bc+cd+da)=$$ $$=1-\frac{1}{12}(a+c)(b+d)\geq1-\frac{\left(\frac{a+b+c+d}{2}\right)^2}{12}=1-\frac{1}{3}=\frac{2}{3}.$$ For $(a,b,c,d)\rightarrow(2,2,0,0)$ we see that $S\rightarrow\frac{2}{3}$,which says that $\frac{2}{3}$ is infimum

and the minimum does not exist.

Done!