Show that $\frac{2a_1^2}{a_1+a_2}+\frac{2a_2^2}{a_2+a_3}+...+\frac{2a_n^2}{a_n+a_1}\geq a_1+a_2+...+a_n$ 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?
 A: Consider the function $f(x)=\frac{2}{1+x}$ for $x>0$. Then $f$ is convex so with $S=\sum_{i=1}^na_i$, we can use Jensen's inequality to infer that
$$
\frac{a_1}{S}f(a_2/a_1)+\cdots+\frac{a_n}{S}f(a_1/a_n)\geq f\left(\frac{a_2}{S}+\frac{a_3}{S}+\cdots+\frac{a_1}{S}\right)=f(1)=1.
$$
Thus,
$$
\sum_{i=1}^na_i=S\leq a_1f(a_2/a_1)+\cdots+a_nf(a_1/a_n)=\frac{2a_1}{1+a_2/a_1}+\cdots+\frac{2a_n}{1+a_1/a_n}
$$
which simplifies to the LHS of your inequality.
A: Here is another way:
$$\frac{2a_k^2}{a_k+a_{k+1}} + \frac{a_k + a_{k+1}}2 \ge 2a_k$$
by AM-GM. Sum this cyclically and simplify  to get your inequality.
A: $$\frac{2a_1^2}{a_1+a_2}+\frac{2a_2^2}{a_2+a_3}+...+\frac{2a_n^2}{a_n+a_1}\geq a_1+a_2+...+a_n\tag{1}$$
Thus to prove (1) hold, it is suffice for us to add $2A(a_1+a_2+...+a_n)$ on both sides and and to prove:
$$\frac{2a_1^2}{a_1+a_2}+A(a_1+a_2)+\frac{2a_2^2}{a_2+a_3}+A(a_2+a_3)+$$
$$+...+\frac{2a_n^2}{a_n+a_1}+A(a_n+a_1)\geq (2A+1)(a_1+a_2+...+a_n).\tag{2}$$
Since $\frac{2a_1^2}{a_1+a_2}+A(a_1+a_2)\geq 2\sqrt{2A}a_1$, we have to prove:
$$2\sqrt{2A}(a_1+a_2+...+a_n)\geq (2A+1)(a_1+a_2+...+a_n)\tag{3}$$
or
$$2\sqrt{2A}\geq 2A+1\tag{4}$$
And we found out that $A=1/2$ is the only solution.
A: 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.
