Ukrajina 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 $\frac{a_1^2}{b_1}+\cdots +\frac{a_n^2}{b_n}\ge \frac{(a_1+\cdots +a_n{)}^2}{b_1+\cdots +b_n}$. 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.