Inegalitate conditionata cu suma

Marius Mainea · Post by **Marius Mainea** » Fri Feb 06, 2009 9:34 pm

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

Arbos · Post by **Arbos** » Sat Mar 14, 2009 9:23 pm

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}) $

alex2008 · Post by **alex2008** » Mon May 04, 2009 10:13 am

$ \sum_{cyc}a(\sqrt{b}+\sqrt{c})=\sum_{cyc}(a-a\sqrt{a})\le \sum_{cyc}\frac{\sqrt{a}}{4}=\frac{1}{4} $

Arbos · Post by **Arbos** » Mon Jan 11, 2010 9:41 am

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} $

Marius Mainea · Post by **Marius Mainea** » Mon Jan 11, 2010 8:35 pm

Notand $ \sqrt{a}=x $si analoagele inegalitatea este echivalenta cu

$ \sum x^2(y+z)+3xyz\le \frac{(x+y+z)^3}{3} $