An algebra problem by Anas Kudsi

WLOG assume that $a+b\ge a+c\ge b+c$ then using Chebyshev have: $\sum_{cyc}\frac{a^{10}}{(b+c) (b^3+c^3)}\ge \frac{1}{3}$ $(\sum_{cyc}\frac{a^6}{b^3+c^3})(\sum_{cyc}\frac{a^4}{b+c})\ge \frac{1}{3}\frac{a^3+b^3+c^3}{2}\frac{(a^2+b^2+c^2)^2}{2(a+b+c)}$ also have $(a+b+c)^2\ge 3(ab+bc+ac)=3$ and by Hölder $9(a^3+b^3+c^3)\ge(a+b+c)^3$ hence $\frac{1}{3}\frac{a^3+b^3+c^3}{2}\frac{(a^2+b^2+c^2)^2}{2(a+b+c)}\ge \frac{1}{3}(\frac{(a+b+c)^3}{18})^2\ge\frac{9}{18^2}=\frac{1}{36}$ So the minimum is $\frac{1}{36}$ Done.