Fie a,b,c numere reale pozitive cu proprietatea:
\( \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1. \)
Sa se arate ca
\( a+b+c \geq ab+bc+ca \).
JBTST III 2007, Problema 4
Moderators: Laurian Filip, Filip Chindea, Radu Titiu, maky, Cosmin Pohoata
-
Laurian Filip
- Site Admin
- Posts: 344
- Joined: Sun Nov 25, 2007 2:34 am
- Location: Bucuresti/Arad
- Contact:
-
Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
\( \sum_{cyc}{} {\frac {1}{a+b+1}} = 3 - \sum_{cyc}{} {\frac {a+b}{a+b+1}} \geq 1 \Rightarrow 2 \geq \sum_{cyc}{} {\frac {a+b}{a+b+1}}= \sum_{cyc}{} {\frac {a^2}{a^2+ab+a}} + \sum_{cyc}{} {\frac {a^2}{a^2+ac+a}} \)
\( \frac {Cauchy}{\geq} \) \( \frac {4({a+b+c})^2}{2 \sum_{cyc}{} {(a^2+ab+a)}} = \frac {2({a+b+c})^2}{\sum_{cyc}{} {(a^2+ab+a)}} \) \( \Rightarrow \sum_{cyc}{} {a^2} + \sum_{cyc}{} {ab} + \sum_{cyc}{} {a} \geq ({a+b+c})^2 \Rightarrow a+b+c \geq ab+bc+ca \) QED
\( \frac {Cauchy}{\geq} \) \( \frac {4({a+b+c})^2}{2 \sum_{cyc}{} {(a^2+ab+a)}} = \frac {2({a+b+c})^2}{\sum_{cyc}{} {(a^2+ab+a)}} \) \( \Rightarrow \sum_{cyc}{} {a^2} + \sum_{cyc}{} {ab} + \sum_{cyc}{} {a} \geq ({a+b+c})^2 \Rightarrow a+b+c \geq ab+bc+ca \) QED
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers