Inegalitate conditionata cu produsul 1

[Claudiu Mindrila]

Fie \( a,b,c \) numere reale strict pozitive astfel incat \( abc=1 \). Demonstrati ca \( \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 \right) ^2 \ge 3(a+b+c)+2(\sqrt{a}+\sqrt{b}+\sqrt{c})+1. \)

[Marius Mainea]

Ridicand la patrat si reducand termenii asemenea obtinem

\( \sum_{cyc}{a^2b^2}+2\sum_{cyc}{ab}\ge \sum_{cyc}a+2\sum_{cyc}{\sqrt{a}} \)

care este evidenta.