Fie \( x_1,x_2,...,x_n\geq0 \) cu \( x_1\cdot...\cdot x_n=1 \). Demonstrati inegalitiatea:
\( \sum x_i^{k+1}\geq\sum x_i^{k} \)
Inegalitate utila....
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Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
\( x_1x_2...x_n=1 \Rightarrow \frac {1}{n} (\sum_{i=1}^{n} {x_i}) \geq 1 \)
Acum avem: \( \sum_{i=1}^{n} {{x_i}^{k+1}} \) \( \geq \) \( \frac {1}{n} (\sum_{i=1}^{n} {x_i})(\sum_{i=1}^{n} {{x_i}^k}) \) \( \geq \) \( \sum_{i=1}^{n} {{x_i}^k} \) QED
Acum avem: \( \sum_{i=1}^{n} {{x_i}^{k+1}} \) \( \geq \) \( \frac {1}{n} (\sum_{i=1}^{n} {x_i})(\sum_{i=1}^{n} {{x_i}^k}) \) \( \geq \) \( \sum_{i=1}^{n} {{x_i}^k} \) QED
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers