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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?

  • 3
    The last inequality can't possibly be true. If $xyz=1$, then $\frac{1}{x}\frac{1}{y}\frac{1}{z}=1$, and the only way the inequality can hold for both $x,y,z$ and $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ is if $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, which is clearly not true in general. You're going to need to start with something sharper than the AM-GM...2012-08-29

2 Answers 2