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 \).
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Mateescu Constantin
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Marius Mainea
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Folosind AM-GM avem :
\( LHS\ge\frac{1}{2}(\sum \frac{x^2}{y^2}+9)=\frac{1}{2}[(\frac{x^2}{y^2}+1)+(\frac{y^2}{z^2}+1+1+1)+(\frac{z^2}{x^2}+1+1+1+1+1)]\ge\frac{x}{y}+2\sqrt{\frac{y}{z}}+3\sqrt[3]{\frac{z}{x}}=\frac{x}{y}+\sqrt{\frac{y}{z}}+\sqrt{\frac{y}{z}}+\sqrt[3]{\frac{z}{x}}+\sqrt[3]{\frac{z}{x}}+\sqrt[3]{\frac{z}{x}}\ge 6\sqrt[6]{\frac{x}{y}\sqrt{\frac{y}{z}}^2\sqrt[3]{\frac{z}{x}}^3}=LHS \)
\( LHS\ge\frac{1}{2}(\sum \frac{x^2}{y^2}+9)=\frac{1}{2}[(\frac{x^2}{y^2}+1)+(\frac{y^2}{z^2}+1+1+1)+(\frac{z^2}{x^2}+1+1+1+1+1)]\ge\frac{x}{y}+2\sqrt{\frac{y}{z}}+3\sqrt[3]{\frac{z}{x}}=\frac{x}{y}+\sqrt{\frac{y}{z}}+\sqrt{\frac{y}{z}}+\sqrt[3]{\frac{z}{x}}+\sqrt[3]{\frac{z}{x}}+\sqrt[3]{\frac{z}{x}}\ge 6\sqrt[6]{\frac{x}{y}\sqrt{\frac{y}{z}}^2\sqrt[3]{\frac{z}{x}}^3}=LHS \)