Find infimum of expression$\sqrt[5]{\frac{a}{b+c}}+\sqrt[5]{\frac{b}{c+d}}+\sqrt[5]{\frac{c}{d+e}}+\sqrt[5]{\frac{d}{e+a}}+\sqrt[5]{\frac{e}{a+b}}$ for $a,b,c,d,e \in \mathbb{R}_+$.
I tried a lot of methods, i.e. AM-GM inequality, Radon inequality, rearrangement, I searched for some refinements of Nesbitt inequality, tried Lagrange Multipliers, but the computations are very messy. I noticed that this inequality is homogenous, but the form of the fractions isn't nice.
I would be very grateful for any hint, sources which deals with this problems, papers, helpful approaches etc. (not solution).
It's not solution, but maybe it will help. $$\sum_{cyc}\sqrt[5]{\frac{a}{b+c}}>\sum_{cyc}\frac{\sqrt[5]{a}}{\sqrt[5]b+\sqrt[5]c}>$$ $$>\sum_{cyc}\frac{\sqrt[5]a}{\sqrt[5]a+\sqrt[5]b+\sqrt[5]c+\sqrt[5]d+\sqrt[5]e}=1.$$ Also, for $c=e\rightarrow0^+$ by AM-GM we obtain: $$\sum_{cyc}\sqrt[5]{\frac{a}{b+c}}\rightarrow\sqrt[5]{\frac{a}{b}}+\sqrt[5]{\frac{b}{d}}+\sqrt[5]{\frac{d}{a}}\geq3,$$ which says that $1\leq M\leq3.$
Also, since $$\sum_{cyc}\sqrt[5]{\frac{a}{b+c}}>\sqrt[5]{\sum_{cyc}\frac{a}{b+c}}=\sqrt[5]{\sum_{cyc}\frac{a^2}{ab+ac}}\geq$$ $$\geq\sqrt[5]{\frac{(a+b+c+d+e)^2}{\sum\limits_{cyc}(ab+ac)}}\geq \sqrt[5]{\frac{5}{2}},$$ we obtain: $$\sqrt[5]{\frac{5}{2}}\leq M\leq3.$$