Problema 2, lista scurta 2009

[alex2008]

Demonstrati ca pentru orice numar real \( x>0 \) si orice intreg \( n\in \mathbb{N}^* \) are loc inegalitatea :

\( \sum_{k=1}^n\frac{\sqrt{2k-1}}{x+k^2}<\sqrt{\frac{n}{x}}. \)

Dan Nedeianu, Drobeta Turnu Severin

[Marius Mainea]

Aplicam C-B-S :

\( LHS\le\sqrt{n\sum {\frac{2k-1}{(x+k^2)^2}}}\le\sqrt{n\sum {(\frac{1}{x+(k-1)^2}-\frac{1}{x+k^2})}}=\sqrt{n(\frac{1}{x}-\frac{1}{x+n^2})}<RHS \)