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a^2+b^2+c^2=3...

Posted: Tue Jun 24, 2008 2:03 pm
by Marius Mainea
Fie a, b, c, x, y,z numere reale pozitive astfel incat \( a^2+b^2+c^2=3. \) Atunci:

\( \frac{x}{a}+\frac{y}{b}+\frac{z}{c}\geq sqrt {xy}+sqrt {yz}+\sqrt {zx} \)

Marius Damian SHL 2006

Posted: Tue Jun 24, 2008 6:05 pm
by Marius Dragoi
\( \sum_{cyc}{} {\frac {x}{a}} \) \( \frac {Cauchy}{\geq} \) \( \frac {{(\sqrt {x} + \sqrt {y} +\sqrt {z} )}^2}{a+b+c} \) \( \geq \) \( \frac {{(\sqrt {x} +\sqrt {y} + \sqrt {z} )}^2}{3} \) \( \geq \) \( \sqrt {xy} +\sqrt {yz} + \sqrt {zx} \) \( \Leftrightarrow \)

\( \Leftrightarrow \) \( \sum_{cyc}{} {{(\sqrt {x} - \sqrt {y})}^2} \) \( \geq 0 \) adevarat