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Inegalitate in 3 variabile conditionata
Posted: Fri Mar 13, 2009 9:39 am
by zeta
Fie \( a,b,c>0 \) ai \( a+b+c=3. \) Atunci \( \frac{a}{2+b^3}+\frac{b}{2+c^3}+\frac{c}{2+b^3}\geq\frac{1}{6}(5+abc) \).
Posted: Fri Mar 13, 2009 5:09 pm
by zeta
Observatie: daca nu ma insel, inegalitatea este adevarata ptr orice \( a,b,c>0,\ a+b+c\in (k,3] \), unde \( k \)este aprox \( 2,1742... \).
Posted: Fri Mar 13, 2009 8:11 pm
by Marius Mainea
Posted: Sun Mar 15, 2009 9:32 pm
by zeta
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 \( \geq\frac{1}{6}(\frac{81S-4S^3}{27}+abc) \).