Inegalitate in 3 variabile conditionata

zeta · Post by **zeta** » Fri Mar 13, 2009 9:39 am

Fie \( a,b,c>0 \) ai \( a+b+c=3. \) Atunci \( \frac{a}{2+b^3}+\frac{b}{2+c^3}+\frac{c}{2+b^3}\geq\frac{1}{6}(5+abc) \).

zeta · Post by **zeta** » Fri Mar 13, 2009 5:09 pm

Observatie: daca nu ma insel, inegalitatea este adevarata ptr orice \( a,b,c>0,\ a+b+c\in (k,3] \), unde \( k \)este aprox \( 2,1742... \).

Marius Mainea · Post by **Marius Mainea** » Fri Mar 13, 2009 8:11 pm

Vezi aici.

zeta · Post by **zeta** » Sun Mar 15, 2009 9:32 pm

Daca notam \( S=a+b+c \), atunci tinand cont ca \( ab^2+bc^2+ca^2\leq\frac{4S^3}{27}-abc \) (din solutia de pe mathlinks) rezulta ca membrul stang al inegalitatii este \( \geq\frac{1}{6}(\frac{81S-4S^3}{27}+abc) \).