# Inequality with summation If ai positive numbers and n>=2 (the subscripts are taken modulo n), how can I prove the following inequality n sum_(k=1)^n 1/((n-1)a_k+a_(k+1))<=sum_(k=1)^n 1/(a_k)?

Inequality with summation
If ${a}_{i}$ positive numbers and $n\ge 2$ (the subscripts are taken modulo $n$), how can I prove the following inequality
$n\sum _{k=1}^{n}\frac{1}{\left(n-1\right){a}_{k}+{a}_{k+1}}\le \sum _{k=1}^{n}\frac{1}{{a}_{k}}$?
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hampiova76
By C-S
$n\sum _{k=1}^{n}\frac{1}{\left(n-1\right){a}_{k}+{a}_{k+1}}\le \frac{n}{\left(n-1+1{\right)}^{2}}\sum _{k=1}^{n}\left(\frac{\left(n-1{\right)}^{2}}{\left(n-1\right){a}_{k}}+\frac{{1}^{2}}{{a}_{k+1}}\right)=$
$=\frac{1}{n}\sum _{k=1}^{n}\left(\frac{n-1}{{a}_{k}}+\frac{1}{{a}_{k+1}}\right)=\frac{1}{n}\sum _{k=1}^{n}\left(\frac{n-1}{{a}_{k}}+\frac{1}{{a}_{k}}\right)=\sum _{k=1}^{n}\frac{1}{{a}_{k}}.$