, 2008

Applying the Cauchy-Bunyakowsky inequality to sets $\frac{a_1}{\sqrt{b_1}}, \dots, \frac{a_n}{\sqrt{b_n}}$ and $\sqrt{b_1}, \dots, \sqrt{b_n}$ in which numbers $a_1, \dots, a_n, b_1, \dots, b_n$ are positive we find that $\frac{a_1^2}{b_1} + \dots + \frac{a_n^2}{b_n} \ge \frac{(a_1+\dots+a_n)^2}{b_1+\dots+b_n}$. Next we rearrange the inequality as follows:
$$\frac{a+b}{2b+c} + \frac{b+c}{2c+a} + \frac{c+a}{2a+b} = \frac{(a+b)^2}{(a+b)(2b+c)} + \frac{(b+c)^2}{(b+c)(2c+a)} + \frac{(c+a)^2}{(c+a)(2a+b)}$$
$\ge$ (we use the mentioned above inequality here)
$$\ge \frac{((a+b)+(b+c)+(c+a))^2}{(a+b)(2b+c)+(b+c)(2c+a)+(c+a)(2a+b)} \ge 2.$$
In order to verify the last rearrangement, simply expand the brackets and summarize similar summands.