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Problema de Marin Chirciu
Posted: Fri May 29, 2009 11:53 pm
by alex2008
1.Fie \( a,b,c \in (0, \infty) \) astfel incat \( abc=\sqrt[3]{2} \) . Sa se demonstreze ca :
\( \frac{1}{a^3(1+b^3)}+\frac{1}{b^3(1+c^3)}+\frac{1}{c^3(1+a^3)} \geq 1 \)
Posted: Mon Jun 01, 2009 10:42 am
by Mateescu Constantin
Sa demonstram mai intai inegalitatea:
\( \frac{1}{x(1+y)}+\frac{1}{y(1+z)}+\frac{1}{z(1+x)}\ge \frac{3}{\sqrt[3]{xyz}(1+\sqrt[3]{xyz}). \)
Avem \( (1+xyz)\left\(\frac{1}{x(1+y)}+\frac{1}{y(1+z)}+\frac{1}{z(1+x)}\right\)+3=\sum\frac{1+xyz+x+xy}{x(1+y)}=\sum\frac{1+x}{x(1+y)}+\sum\frac{y(z+1)}{1+y} \)
Din \( AM-GM \) \( \Longrightarrow \sum\frac{1+x}{x(1+y)}+\sum\frac{y(z+1)}{1+y} \ge \frac{3}{\sqrt[3]{xyz}}+3\sqrt[3]{xyz} \)
Ramane sa aratam \( \frac{\frac{3}{\sqrt[3]{xyz}}+3\sqrt[3]{xyz}-3}{1+xyz}\ge \frac{3}{\sqrt[3]{xyz}(1+\sqrt[3]{xyz})} \) care este chiar o egalitate.
Luand \( x=a^3,\ y=b^3,\ z=c^3 \)
\( \Longrightarrow \frac{1}{a^3(1+b^3)}+\frac{1}{b^3(1+c^3)}+\frac{1}{c^3(1+a^3)}\ge \frac{3}{abc(1+abc)}=\frac{3}{\sqrt[3]{2}+\sqrt[3]{4}}>1 \), deci egalitatea cu \( 1 \) nu poate avea loc.
De fapt pentru \( abc\in \(0,\ t\] \ LHS\ \ge\ 1 \), unde \( t \) este radacina pozitiva a ecuatiei \( t^2+t-3=0 \) (Egalitate pentru \( abc=t \)).