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Shortlist 9

Posted: Sun Mar 28, 2010 7:52 pm
by Antonache Emanuel
Fie a,b,c numere strict pozitive, subunitare cu suma 1. Demonstrati ca:
\( \sum\frac{log^2_{ab}c}{a+b}\ge\frac{9}{8} \).
Tudorel Lupu, Constanta, Shortlist 2003

Posted: Sun Mar 28, 2010 7:59 pm
by Marius Mainea
\( LHS\ge \frac{(\sum\log_{ab}c)^2}{2\sum a}\ge RHS \)