Inegalitate cu radicali de ordinul 8
Posted: Wed Jun 10, 2009 5:40 pm
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
\( \sqrt[8]{\frac{x+y+z}{3}}\ge \frac{\sqrt[8]{x}+\sqrt[8]{y}+\sqrt[8]{z}}{3} \).
Catalin Cristea