I would like to show that:
$ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3 $
Using AM-GM inequality:
$ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3\frac{(a+2b)(b+2c)(c+2a)}{(a+2c)(b+2a)(c+2b)} $
It suffices to show that:
$ \frac{(a+2b)(b+2c)(c+2a)}{(a+2c)(b+2a)(c+2b)} \geq1 $
$ \Longleftrightarrow ab^2+bc^2+ca^2-(a^2b+b^2c+c^2a)\geq0$
$ \Longleftrightarrow x+y+z\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
$ xyz=1$
($x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$)
How can I prove this last inequality? Is there any simpler proof?