parmenides51 wrote: For positive real numbers $a, b$ and $c$, prove that $$\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{ c^2+ca+a^2}\ge\frac{a+b+c}{3}$$ https://artofproblemsolving.com/community/c6h381861p2114521 https://artofproblemsolving.com/community/c6h1272771p6659190 https://artofproblemsolving.com/community/c6h589771p3496359 Prove that if $a, b, c\ge 0$ and $ab+bc+ca>0$ then $$\frac{a^2+b^2+c^2}{a+b+c}\leq\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\leq\frac{3(a^3+b^3+c^3)}{(a+b+c)^2}$$For positive real numbers $a, b$ and $c$, prove that $$\frac{{{a}^{3}}}{6{{a}^{2}}+ab+{{b}^{2}}}+\frac{{{b}^{3}}}{6{{b}^{2}}+bc+{{c}^{2}}}+\frac{{{c}^{3}}}{6{{c}^{2}}+ca+{{a}^{2}}}\ge \frac{a+b+c}{8}$$

solved also here (as p2)

Let $a, b, c\ge 0$ and $ab+bc+ca>0.$ Prove that $$\frac{a^3}{a^2-ab+b^2}+\frac{b^3}{b^2-bc+c^2}+\frac{c^3}{c^2-ca+a^2}\leq\frac{9(a^3+b^3+c^3)}{(a+b+c)^2}$$

$\sum{\frac{a^3}{a^2+ab+b^2}} \geq \frac{a+b+c}{3}$(?) $\sum{\frac{a^3}{a^2+ab+b^2}} -a \geq \frac{-2(a+b+c)}{3}$(?) $\sum{\frac{ab(a+b)}{a^2+ab+b^2}} \geq \frac{2(a+b+c)}{3}$(?) $AM-GM$ for left $\sum{\frac{ab(a+b)}{3ab}} \geq \frac{2(a+b+c)}{3}$(?) $\sum{\frac{(a+b)}{3}} \geq \frac{2(a+b+c)}{3}$ (•) We are done