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Iasi 2006
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Iasi 2006
Posted:
Sun Feb 01, 2009 1:01 pm
by
alex2008
Fie
\( x,y,z\in \mathbb{R} \)
cu
\( xyz=1 \)
. Sa se arate ca:
\( \sum_{cyc}\frac{1+xy}{1+z}\ge 3 \)
.
Etapa locala Iasi 2006
Posted:
Sun Feb 01, 2009 1:16 pm
by
Claudiu Mindrila
Avem:
\( \sum\frac{1+xy}{1+z}=\sum\frac{1+xy}{z\left(1+xy\right)}=\sum\frac{1}{z}=\sum xy\ge3\sqrt[3]{\left(xyz\right)^{2}}=3 \)
.