O inegalitate interesanta

[Claudiu Mindrila]

Daca \( a_{i}\ge0,\ b_{i}>0,\ i=\overline{1,\ k},\ k,\ n\in\mathbb{N},\ k\ge1 \) atunci \( \frac{a_{1}^{n+1}}{b_{1}^{n}}+\frac{a_{2}^{n+1}}{b_{2}^{n}}+\dots+\frac{a_{k}^{n+1}}{b_{k}^{n}}\ge\frac{a_{1}+a_{2}+\dots+a_{k}}{b_{1}+b_{2}+\dots+b_{k}}\left(\frac{a_{1}^{n}}{b_{1}^{n-1}}+\frac{a_{2}^{n}}{b_{2}^{n-1}}+\dots+\frac{a_{k}^{n}}{b_{k}^{n-1}}\right)\ \).

[Mateescu Constantin]

Vezi atasamentul de aici .