Page 1 of 1
Inegalitate conditionata trei variabile
Posted: Mon Jan 28, 2008 6:55 pm
by Tudorel Lupu
Fie \( a, b, c\in (-1, \infty) \) astfel incat sa avem \( ab+bc+ca+2abc=1 \). Sa se arate ca
\( \frac{1}{2+a+b}+\frac{1}{2+b+c}+\frac{1}{2+c+a}\leq 1 \)
Tudorel Lupu, Olimpiada locala Constanta, 2008
Posted: Tue Jan 29, 2008 9:40 am
by mihai++
Din \( HM\leq AM \) avem \( \frac{1}{2+a+b}\leq \frac{\frac{1}{1+a}+\frac{1}{1+b}}{4} \) si analoagele.
In consecinta \( \sum \frac{1}{2+a+b} \leq \frac{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}{2} \) si prin calcul, obtinem ca \( \frac{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}{2}=1\leftrightarrow ab+bc+ca+2abc=1 \).
Egalitatea se obtine atunci cand \( 1+a=1+b=1+c \rightarrow a=b=c \rightarrow 2(a+1)^2(a-\frac{1}{2})=0 \rightarrow a=b=c=\frac{1}{2} \)
Posted: Fri Feb 01, 2008 12:22 am
by mumble
Solutie poate surprinzatoare prin "straightforward". Dupa ce aducem la acelasi numitor ineg e echiv cu:
\( \ 12+8\sum_{cyc}a+3\sum_{cyc}ab+\sum_{cyc}a^2\leq8+2\sum_{cyc}a^2+8\sum_{cyc}a+6\sum_{cyc}ab+2abc+\sum_{cyc}ab(a+b)\Leftrightarrow\\4\leq1+2\sum_{cyc}ab+\sum_{cyc}a^2+\sum_{cyc}ab(a+b). \)
Din ineg mediilor:
\( RHS\geq3ab+3bc+3ca+6abc+1=4 \)