Dubla inegalitate
Posted: Tue Jun 02, 2009 9:07 pm
Daca \( x,\ y,\ z\ >\ 0 \), demonstrati ca:
\( \frac{3}{2}\ \left\(\frac{x^2}{y^2}\ +\ \frac{y^2}{z^2}\ +\ \frac{z^2}{x^2}\ +\ 1\right\)\ \ge\ \frac{x}{y}\ +\ 2\sqrt{\frac{y}{z}}+\ 3\sqrt[3]{\frac{z}{x}}\ \ge \ 6 \).
\( \frac{3}{2}\ \left\(\frac{x^2}{y^2}\ +\ \frac{y^2}{z^2}\ +\ \frac{z^2}{x^2}\ +\ 1\right\)\ \ge\ \frac{x}{y}\ +\ 2\sqrt{\frac{y}{z}}+\ 3\sqrt[3]{\frac{z}{x}}\ \ge \ 6 \).