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Posted: **Sun Feb 22, 2009 6:55 pm**

Fie \( a,b,c>0 \) cu \( a+b+c=1 \). Atunci are loc inegalitatea:
\( \frac{ab}{1+c}+\frac{bc}{1+a}+\frac{ca}{1+b}\leq \frac{1}{4} \).

V. Cartoaje

Posted: **Sun Feb 22, 2009 8:44 pm**

\( LHS=\sum{\frac{ab}{(c+a)+(c+b)}}\le \sum{\frac{ab}{4}\(\frac{1}{c+a}+\frac{1}{c+b}\)}=\frac{1}{4}\sum{\frac{ab+bc}{a+c}}=\frac{1}{4}\sum b=RHS \)

Inegalitate V. Cartoaje