Galati 2008

[alex2008]

Sa se arate ca oricare ar fi numerele reale \( a,b,c \in(0;1) \) , avem inegalitatea :
\( \sqrt{\frac{ab}{ab+c\cdot(a+b+c)}}+\sqrt{\frac{bc}{bc+a\cdot(a+b+c)}}+\sqrt{\frac{ca}{ca+b\cdot(a+b+c)}}\le \frac{3}{2} \)

[Marius Mainea]

Folosind inegalitatea mediilor avem :
\( \sum\sqrt{\frac{ab}{ab+c(a+b+c)}}=\sum{sqrt{\frac{ab}{(c+a)(c+b)}}}\le \sum{\frac{1}{2}(\frac{a}{c+a}+\frac{b}{c+b})}=\frac{1}{2}(1+1+1)=\frac{3}{2} \)