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Din nou inegalitate

Posted: Tue Mar 31, 2009 8:05 pm
by Claudiu Mindrila
Fie \( a,b,c \) numere reale pozitive. Demonstrati ca \( \frac{ab}{4a+3b+2c}+\frac{bc}{4b+3c+2a}+\frac{ca}{4c+3a+2b}\le\frac{a+b+c}{9} \) .

Baleanu Andrei Razvan, Mathematical Reflections 2/2008

Posted: Wed Apr 01, 2009 9:53 pm
by Marius Mainea
Se poate demonstra o inegalitate mai generala:

\( \frac{ab}{xa+yb+2c}+\frac{bc}{xb+yc+2a}+\frac{ca}{xc+ya+2b}\le \frac{a+b+c}{x+y+2} \) pentru orice a,b,c pozitive si \( x,y\ge 1 \)

Pentru \( x=\frac{6}{5} \) si \( y=\frac{8}{5} \) se obtine problema 5 propusa de Nikolai Nicolov la TST 47. IMO Sofia , 2006.