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Inegalitate conditionata cu suma
Posted: Fri Feb 06, 2009 9:34 pm
by Marius Mainea
Daca a, b, c sunt numere reale nenegative cu \( \sqrt{a}+\sqrt{b}+\sqrt{c}=1 \), atunci :
\( a(\sqrt{b}+\sqrt{c})+b(\sqrt{c}+\sqrt{a})+c(\sqrt{a}+\sqrt{b})\le \frac{1}{4} \)
G.M. seria B
Inegalitate conditionata
Posted: Sat Mar 14, 2009 9:23 pm
by Arbos
Obs. Luand a=0 obtinem urmatorul tip de inegalitate conditionata:
"Daca \( x+y=1 \) atunci \( $x^2y+y^2x \leq\frac{1}{4}$ \)", adevarata din \( xy\leq\frac{(x+y)^2}{4} \).
Fie acum \( x=\sqrt{a}+sqrt{b} \) si \( y=\sqrt{c} \) cu alegerea lui c cel mai mare dintre a si b. Inegalitatea din Obs. devine: \( (\sqrt{a}+\sqrt{b})^2\sqrt{c}+c(\sqrt{a}+\sqrt{b})\leq\frac{1}{4} \). Se arata usor ca \( a(\sqrt{b}+\sqrt{c})+b(\sqrt{c}+\sqrt{a})+c(\sqrt{a}+\sqrt{b})\leq\(\sqrt{a}+\sqrt{b})^2\sqrt{c}+c(\sqrt{a}+\sqrt{b}) \)
Posted: Mon May 04, 2009 10:13 am
by alex2008
\( \sum_{cyc}a(\sqrt{b}+\sqrt{c})=\sum_{cyc}(a-a\sqrt{a})\le \sum_{cyc}\frac{\sqrt{a}}{4}=\frac{1}{4} \)
Inegalitate conditionata cu suma
Posted: Mon Jan 11, 2010 9:41 am
by Arbos
Daca \( sqrt{a}+sqrt{b}+sqrt{c}=1 \), atunci
\( a(\sqrt{b}+\sqrt{c})+b(\sqrt{c}+\sqrt{a})+c(\sqrt{a}+\sqrt{b})+3sqrt{abc}\leq\frac{1}{3} \)
Posted: Mon Jan 11, 2010 8:35 pm
by Marius Mainea
Notand \( \sqrt{a}=x \)si analoagele inegalitatea este echivalenta cu
\( \sum x^2(y+z)+3xyz\le \frac{(x+y+z)^3}{3} \)