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Fie a,b,c numere reale pozitive cu proprietatea:

\( \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1. \)

Sa se arate ca

\( a+b+c \geq ab+bc+ca \).

\( \sum_{cyc}{} {\frac {1}{a+b+1}} = 3 - \sum_{cyc}{} {\frac {a+b}{a+b+1}} \geq 1 \Rightarrow 2 \geq \sum_{cyc}{} {\frac {a+b}{a+b+1}}= \sum_{cyc}{} {\frac {a^2}{a^2+ab+a}} + \sum_{cyc}{} {\frac {a^2}{a^2+ac+a}} \)

\( \frac {Cauchy}{\geq} \) \( \frac {4({a+b+c})^2}{2 \sum_{cyc}{} {(a^2+ab+a)}} = \frac {2({a+b+c})^2}{\sum_{cyc}{} {(a^2+ab+a)}} \) \( \Rightarrow \sum_{cyc}{} {a^2} + \sum_{cyc}{} {ab} + \sum_{cyc}{} {a} \geq ({a+b+c})^2 \Rightarrow a+b+c \geq ab+bc+ca \) QED