Inequality with minimal value

[MM]

Determine the least value of
[tex]\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}[/tex]
over all positive real numbers a, b, c.

[Rock'n'roller]

[tex]\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} = \frac{a^2}{ab+2ca}+\frac{b^2}{bc+2ab}+\frac{c^2}{ca+2bc} \ge \frac{(a+b+c)^2}{3(ab+bc+ca)} \ge 1 ( \Leftrightarrow (a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0 )[/tex], so the least value is 1.