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
Inegalitate conditionata trei variabile
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Tudorel Lupu
- Euclid
- Posts: 15
- Joined: Mon Oct 01, 2007 8:58 pm
- Location: Constanta
Inegalitate conditionata trei variabile
Last edited by Tudorel Lupu on Fri Feb 01, 2008 11:13 am, edited 1 time in total.
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 \( 1+a=1+b=1+c \rightarrow a=b=c \rightarrow 2(a+1)^2(a-\frac{1}{2})=0 \rightarrow a=b=c=\frac{1}{2} \)
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 \( 1+a=1+b=1+c \rightarrow a=b=c \rightarrow 2(a+1)^2(a-\frac{1}{2})=0 \rightarrow a=b=c=\frac{1}{2} \)
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:
\( RHS\geq3ab+3bc+3ca+6abc+1=4 \)
\( \ 12+8\sum_{cyc}a+3\sum_{cyc}ab+\sum_{cyc}a^2\leq8+2\sum_{cyc}a^2+8\sum_{cyc}a+6\sum_{cyc}ab+2abc+\sum_{cyc}ab(a+b)\Leftrightarrow\\4\leq1+2\sum_{cyc}ab+\sum_{cyc}a^2+\sum_{cyc}ab(a+b). \)
Din ineg mediilor:
\( RHS\geq3ab+3bc+3ca+6abc+1=4 \)