Cho x, y, z là 3 số thực dương thỏa mãn \...

C

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Ta có:

\(\begin{array}{l}A = \frac{2}{{{x^2} + {y^2}}} + \frac{2}{{{y^2} + {z^2}}} + \frac{2}{{{z^2} + {x^2}}} - \frac{{{x^3} + {y^3} + {z^3}}}{{2xyz}} = \frac{2}{{{x^2} + {y^2}}} + \frac{2}{{{y^2} + {z^2}}} + \frac{2}{{{z^2} + {x^2}}} - \frac{{{x^2}}}{{2yz}} - \frac{{{y^2}}}{{2xz}} - \frac{{{z^2}}}{{2xy}}\\ = \left( {\frac{2}{{{x^2} + {y^2}}} - \frac{{{z^2}}}{{2xy}}} \right) + \left( {\frac{2}{{{y^2} + {z^2}}} - \frac{{{x^2}}}{{2yz}}} \right) + \left( {\frac{2}{{{z^2} + {x^2}}} - \frac{{{y^2}}}{{2xz}}} \right)\end{array}\)

Áp dụng BĐT Cô si, ta có:

\(0 < 2xy \le {x^2} + {y^2},\,\,(x,\,y > 0) \Rightarrow \frac{{{z^2}}}{{2xy}} \ge \frac{{{z^2}}}{{{x^2} + {y^2}}} \Rightarrow \frac{2}{{{x^2} + {y^2}}} - \frac{{{z^2}}}{{2xy}} \le \frac{2}{{{x^2} + {y^2}}} - \frac{{{z^2}}}{{{x^2} + {y^2}}} = \frac{{2 - {z^2}}}{{{x^2} + {y^2}}} = \frac{{{x^2} + {y^2}}}{{{x^2} + {y^2}}} = 1\)

Tương tự, \(\frac{2}{{{y^2} + {z^2}}} - \frac{{{x^2}}}{{2yz}} \le 1\), \(\frac{2}{{{z^2} + {x^2}}} - \frac{{{y^2}}}{{2xz}} \le 1\)

\(\begin{array}{l} \Rightarrow A = \left( {\frac{2}{{{x^2} + {y^2}}} - \frac{{{z^2}}}{{2xy}}} \right) + \left( {\frac{2}{{{y^2} + {z^2}}} - \frac{{{x^2}}}{{2yz}}} \right) + \left( {\frac{2}{{{z^2} + {x^2}}} - \frac{{{y^2}}}{{2xz}}} \right) \le 1 + 1 + 1 = 3\\ \Rightarrow MaxA = 3 \Leftrightarrow \left\{ \begin{array}{l}x = y = z\\{x^2} + {y^2} + {z^2} = 2\end{array} \right. \Leftrightarrow x = y = z = \sqrt {\frac{2}{3}} \end{array}\)