Aratati ca \( (\forall)n\geq 2 \), \( n\in\mathbb{N} \) are loc inegalitatea:
\( \lfloor \sqrt{1}\rfloor^2+\lfloor\sqrt{2}\rfloor^2+\dots+\lfloor\sqrt{n^2-1}\rfloor^2\geq\frac{n^2\cdot(n-1)\cdot(4n+1)^2}{36(n+1)} \)
Prof. Carmen Botea, Braila
Concursul Trident Braila 2009 subiectul 2
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Suma din stanga se poate calcula in functie de \( n \). Dupa aia e simpla problema... 
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