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Posted: **Sun Feb 01, 2009 1:01 pm**

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**

Avem: \( \sum\frac{1+xy}{1+z}=\sum\frac{1+xy}{z\left(1+xy\right)}=\sum\frac{1}{z}=\sum xy\ge3\sqrt[3]{\left(xyz\right)^{2}}=3 \).