Inequality with summation

Question

If $a_i$ positive numbers and $n\ge2$ (the subscripts are taken modulo $n$), how can I prove the following inequality $n\sum\limits_{k=1}^{n} \frac{1}{(n-1)a_k+a_{k+1}}\le\sum\limits_{k=1}^{n} \frac{1}{a_k}$ ?

Answer 1

By C-S $$n\sum_{k=1}^n\frac{1}{(n-1)a_k+a_{k+1}}\leq\frac{n}{(n-1+1)^2}\sum_{k=1}^n\left(\frac{(n-1)^2}{(n-1)a_k}+\frac{1^2}{a_{k+1}}\right)=$$ $$=\frac{1}{n}\sum_{k=1}^n\left(\frac{n-1}{a_k}+\frac{1}{a_{k+1}}\right)=\frac{1}{n}\sum_{k=1}^n\left(\frac{n-1}{a_k}+\frac{1}{a_k}\right)=\sum_{k=1}^n\frac{1}{a_k}.$$