Q:

Is it a telescopic inequality??

Let $a_{i}>0, i=1,2,3,...,n. a_{n+1}=a_{1}$. Prove that $$2\sum_{k=1}^n \frac {a_{k}^2}{a_{k}+a_{k+1}} \ge \sum_{k=1}^n a_k$$ I took the RHS to the LHS and simplified which gave me the following expression after which i don't know what to do :- $$\sum_{k=1}^n a_k\left(\frac {a_k-a_{k+1}}{a_k+a_{k+1}}\right)\ge 0$$

A:

Put
$$S=\sum_{k=1}^n a_k\frac{a_k-a_{k+1}}{a_k+a_{k+1}}$$
and
$$T=\sum_{k=1}^n a_{k+1}\frac{a_k-a_{k+1}}{a_k+a_{k+1}}$$
Then:
$$S-T=\sum_{k=1}^n \frac{(a_k-a_{k+1})^2}{a_k+a_{k+1}}\geq 0$$
$$S+T=\sum(a_k-a_{k+1})=0$$
as $a_{n+1}=a_1$.
Now if we add, we get $2S\geq 0$ and we are done. 

