parmenides51 wrote: Let $a, b, c$ be positive real numbers. Prove that $$\frac{a^3}{a^2 + bc}+\frac{b^3}{b^2 + ca}+\frac{c^3}{c^2 + ab} \ge \frac{(a^2 + b^2 + c^2)(ab + bc + ca)}{a^3 + b^3 + c^3 + 3abc}$$ https://artofproblemsolving.com/community/c6h2073586p14865987 Let $ a,b,c $ be three positive real numbers . Prove that $$\frac{a^3}{a^2+bc}+\frac{b^3}{b^2+ca}+\frac{c^3}{c^2+ab} \ge \frac{2(a^2+b^2+c^2)+ab+bc+ca}{2(a+b+c)}.$$

parmenides51 wrote: Let $a, b, c$ be positive real numbers. Prove that $$\frac{a^3}{a^2 + bc}+\frac{b^3}{b^2 + ca}+\frac{c^3}{c^2 + ab} \ge \frac{(a^2 + b^2 + c^2)(ab + bc + ca)}{a^3 + b^3 + c^3 + 3abc}$$ By AM-GM inequality and by Angel form of Cauchy-Schwarz's inequaity, we have, $$\sum_{cyc}\left(\frac{a^3}{a^2 + bc}\right)=\sum_{cyc}\left(\frac{a^4}{a^3 + abc}\right)\geqslant \frac{(a^2+b^2+c^2)^2}{\sum_{cyc}a^3+3abc}\geqslant \frac{(a^2+b^2+c^2)(ab+bc+ca)}{\sum_{cyc}a^3+3abc}.$$$\blacksquare$ thank you, best regards !