Inegalitate usurica

[Claudiu Mindrila]

Fie numerele reale \( a_{1},\ a_{2},\ldots,\ a_{n}>1 \) ce satisfac relatia \( \sum_{i=1}^{n}\frac{1}{a_{i}^{2}-1}=1 \). Demonstrati ca \( \sum_{i=1}^{n}\frac{1}{a_{i}+1}\le\frac{n}{\sqrt{n+1}+1} \).

Andrei Razvan Baleanu

[Marius Mainea]

\( LHS^2=\(\sum\sqrt{\frac{a_i-1}{a_i+1}}\cdot\frac{1}{\sqrt{a_i^2-1}}\)^2\le\(\sum\frac{a_i-1}{a_i+1}\)\(\sum\frac{1}{a_i^2+1}\)=n-2\cdot LHS \)