Inegalitate tip ONM
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Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
Inegalitate tip ONM
Fie \( a,b,c >0 \) astfel incat \( \sum_{ciclic}^{} {\frac {1}{a}} \geq \sum_{ciclic}^{} {\frac {a}{b}}. \) Demonstrati ca \( \sum_{ciclic}^{} {\frac {a^2}{c}} \geq \sum_{ciclic}^{} {ab}. \)
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers
Aplicam doar Cauchy:
\( \sum\frac{1}{a}\sum\frac{1}{bc}\geq\sum\frac{b}{c}\sum\frac{1}{bc}\geq(\sum\frac{1}{a})^2 \) din ipoteza si Cauchy. Deci \( \sum\frac{1}{bc}\geq \sum\frac{1}{a}\rightarrow \sum a\geq \sum ab \)
Tot din Cauchy avem ca:
\( \sum a \sum \frac{a^2}{c}\geq (\sum a)^2 \) deci \( \sum \frac{a^2}{c}\geq\sum a \) si acum din cele 2 inegalitati concluzie e evidenta.
Egalitatea se intampla cand \( a=b=c=1 \).
\( \sum\frac{1}{a}\sum\frac{1}{bc}\geq\sum\frac{b}{c}\sum\frac{1}{bc}\geq(\sum\frac{1}{a})^2 \) din ipoteza si Cauchy. Deci \( \sum\frac{1}{bc}\geq \sum\frac{1}{a}\rightarrow \sum a\geq \sum ab \)
Tot din Cauchy avem ca:
\( \sum a \sum \frac{a^2}{c}\geq (\sum a)^2 \) deci \( \sum \frac{a^2}{c}\geq\sum a \) si acum din cele 2 inegalitati concluzie e evidenta.
Egalitatea se intampla cand \( a=b=c=1 \).
n-ar fi rau sa fie bine 