Inegalitate conditionata cu produs
Posted: Mon Mar 02, 2009 9:18 am
Daca \( a,b,c > 0 \) astfel incat \( abc = 1 \) aratati ca :
\( \frac {a(1 + a + ab)}{a + 2b} + \frac {b(1 + b + bc)}{b + 2c} + \frac {c(1 + c + ca)}{c + 2a}\geq\ 3 \)
\( \frac {a(1 + a + ab)}{a + 2b} + \frac {b(1 + b + bc)}{b + 2c} + \frac {c(1 + c + ca)}{c + 2a}\geq\ 3 \)