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Inegalitate cu radicali de ordinul 8

Posted: Wed Jun 10, 2009 5:40 pm
by Mateescu Constantin
Sa se demonstreze ca daca \( x,\ y,\ z\in(0,\ \infty) \) atunci are loc inegalitatea
\( \sqrt[8]{\frac{x+y+z}{3}}\ge \frac{\sqrt[8]{x}+\sqrt[8]{y}+\sqrt[8]{z}}{3} \).

Catalin Cristea

Posted: Thu Jun 11, 2009 12:27 am
by Marius Mainea
Se foloseste Power Mean Inequality:

Functia

\( M_r=\(\frac{a_1^r+a_2^r+...+a_n^r}{n}\)^{\frac{1}{r}} \) \( r\neq 0 \) este strict crescatoare.