TITLE: Show that $\frac{2a_1^2}{a_1+a_2}+\frac{2a_2^2}{a_2+a_3}+...+\frac{2a_n^2}{a_n+a_1}\geq a_1+a_2+...+a_n$
QUESTION [9 upvotes]: Showing that $ \frac{2a_1^2}{a_1+a_2}+\frac{2a_2^2}{a_2+a_3}+...+\frac{2a_n^2}{a_n+a_1}\geq a_1+a_2+...+a_n$ holds for positive $a_i$s.
I've tried adding $a_1+a_2, a_2+a_3,...,a_n+a_1$ respectively to the terms of the LHS and using AM-GM on them, but it didn't get me to the needed point. Also tried AM GM on the whole LHS, but couldn't get much. How is it done?

REPLY [16 votes]: You could also use the Cauchy-Schwarz Inequality (http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality). Note that $$\left(\frac{a_1+a_2}{2}+\frac{a_2+a_3}{2}+...+\frac{a_n+a_1}{2}\right)\left(\frac{2a_1^2}{a_1+a_2}+...+\frac{2a_n^2}{a_n+a_1}\right)\ge (a_1+a_2+...+a_n)^2.$$Dividing both sides by $a_1+a_2+...+a_n$ gives the desired conclusion.