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Inegalitate de la ONM 2002

Posted: Fri Jul 25, 2008 1:05 pm
by Claudiu Mindrila
Fie \( a,b,c \) numere reale strict pozitive, astfel incat \( ab+bc+ca=1 \). Sa se demonstreze ca: \( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \geq \sqrt{3}+\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}. \)

Dinu Teodorescu, ONM 2002

Posted: Sat Jul 26, 2008 11:29 pm
by mumble
Folosind conditia \( ab+bc+ca=1 \) rescriem inegalitatea ca \( \sum \frac{ab+bc+ca}{a+b}\geq \sqrt{3}+\sum \frac{ab}{a+b}\Leftrightarrow\sum\frac{c(a+b)}{a+b}\geq\sqrt{3}\Leftrightarrow a+b+c\geq\sqrt{3}, \) inegalitate care rezulta imediat din \( (a+b+c)^2\geq3\sum ab=3. \)