Quidditch wrote: Let $a, b, c\in (0, 1)$ with $a + b + c = 1$. Prove that $$\frac{a^5+b^5}{a^3+b^3}+\frac{b^5+c^5}{b^3+c^3}+\frac{c^5+a^5}{c^3+a^3}\geq\frac{a}{8+b^3+c^3}+\frac{b}{8+c^3+a^3}+\frac{c}{8+a^3+b^3}.$$ Because $$\sum_{cyc}\frac{a}{8+b^3+c^3}\leq\sum_{cyc}\frac{a}{8+3b-2+3c-2}=\sum_{cyc}\frac{a}{4+3b+3c}\leq$$$$\leq\sum_{cyc}\frac{a}{3a+3b+3c}=\frac{1}{3}=\frac{(a+b+c)^2}{3}\leq\sum_{cyc}a^2=\sum_{cyc}\frac{a^2+b^2}{2}\leq\sum_{cyc}\frac{a^5+b^5}{a^3+b^3}.$$