If $27 + abc = (a +b +c)(ab + ac+ bc)$ so $\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a }\geq \frac{3}{2}$

Question

Let $a$, $b$ and $c$ be positive numbers such that $27 + abc = (a +b +c)(ab + ac+ bc)$. Prove that: $$\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a }\geq \frac{3}{2}$$

I have that $\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a} \geq \frac{a+b+c}{3}$ then if I proof that $a+b+c \geq \frac{9}{2}$ I finish. Also $27 + abc = (a +b +c)(ab + ac+ bc)$ can be factorized to $27=(a+b)(ab + ac + bc + c^2)$.

Answer 1

By Holder $\sum\limits_{cyc}\frac{a^2}{a+2b}=\sum\limits_{cyc}\frac{a^3}{a^2+2ab}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(a^2+2ab)}=\frac{a+b+c}{3}$.

The condition gives $27=(a+b)(a+c)(b+c)$.

Thus, it remains to prove that $\frac{a+b+c}{3}\geq\frac{1}{2}\sqrt[3]{(a+b)(a+c)(b+c)}$ or $\frac{a+b+a+c+b+c}{3}\geq\sqrt[3]{(a+b)(a+c)(b+c)}$, which is AM-GM.

Done!

Answer 2

Michael Rozenberg's answer is nice; just would like to point out that the inequality $\sum\limits_{cyclic} \frac{a^{2}}{a+2b}\geq \frac{(a+b+c)}{3}$ follows from the Cauchy Schwarz inequality directly.