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Sa se demonstreze inegalitatea: \( \frac{ac}{2ac+b^2}+\frac{bc}{2bc+a^2}+\frac{ab}{2ab+c^2}\leq 1 \), oricare ar fi \( a,b,c \) numere reale pozitive.

Nicolae Staniloiu, RMCS 20

\( \sum_{cyc}{} {\frac {2ac}{2ac+b^2}} \leq 2 \Leftrightarrow \)\( \sum_{cyc}{} {\frac {b^2}{2ac+b^2}} \geq 1 \)

dar \( \sum_{cyc}{} {\frac {b^2}{2ac+b^2}} \) \( \frac {Cauchy}{\geq} \) \( \frac {({a+b+c)}^2}{\sum_{cyc}{} {(a^2+2ab)}} \) \( = \) \( \frac {{(a+b+c)}^2}{{(a+b+c)}^2} =1 \) Q.E.D.