Din nou inegalitate
Posted: Tue Mar 31, 2009 8:05 pm
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
Baleanu Andrei Razvan, Mathematical Reflections 2/2008