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O inegalitate interesanta

Posted: Mon Dec 21, 2009 1:57 pm
by Claudiu Mindrila
Daca \( a_{i}\ge0,\ b_{i}>0,\ i=\overline{1,\ k},\ k,\ n\in\mathbb{N},\ k\ge1 \) atunci \( \frac{a_{1}^{n+1}}{b_{1}^{n}}+\frac{a_{2}^{n+1}}{b_{2}^{n}}+\dots+\frac{a_{k}^{n+1}}{b_{k}^{n}}\ge\frac{a_{1}+a_{2}+\dots+a_{k}}{b_{1}+b_{2}+\dots+b_{k}}\left(\frac{a_{1}^{n}}{b_{1}^{n-1}}+\frac{a_{2}^{n}}{b_{2}^{n-1}}+\dots+\frac{a_{k}^{n}}{b_{k}^{n-1}}\right)\ \).

Posted: Mon Dec 21, 2009 3:05 pm
by Mateescu Constantin
Vezi atasamentul de aici .