Page 1 of 1

Inegalitate exponentiala conditionata

Posted: Sat Jun 07, 2008 12:43 pm
by Andrei Velicu
Daca \( a, b>0 \) si \( a+b=1 \), aratati ca \( a^a\cdot b^b\geq \frac{1}{2} \).

Marius Cavachi

Posted: Sat Jun 07, 2008 8:30 pm
by Marius Mainea
Indicatie solutie ,,sofisticata'':

Se deconditioneaza inegalitatea (\( a=\frac{x}{x+y},b=\frac{y}{x+y}) \) si apoi se foloseste faptul ca f(x)=xlnx este convexa.