\(\left( P \right)\) qua \(M\left( {4;1;2} \right)...

B

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

* Giả sử \(A\left( {a;0;0} \right),\,\,B\left( {0;b;0} \right),\,\,C\left( {0;0;c} \right)\) \( \Rightarrow Pt\left( P \right):\,\,\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\).

* \(M\left( {4;1;2} \right) \in \left( P \right) \Rightarrow \dfrac{4}{a} + \dfrac{1}{b} + \dfrac{2}{c} = 1\,\,\left( 1 \right)\)

* \(T = \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}} + \dfrac{1}{{O{C^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{1}{{{c^2}}}\)

* Áp dụng BĐT Bunhiacopxki ta có:

\(\begin{array}{l}\,\,\,\,\,{\left( {\dfrac{1}{a}.4 + \dfrac{1}{b}.1 + \dfrac{1}{c}.2} \right)^2} \le \left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{1}{{{c^2}}}} \right)\left( {{4^2} + {1^2} + {2^2}} \right)\\ \Rightarrow 1 \le T.21 \Rightarrow T \ge \dfrac{1}{{21}} \Rightarrow {T_{\min }} = \dfrac{1}{{21}}\end{array}\)

* Dấu "=" xảy ra \( \Leftrightarrow \left\{ \begin{array}{l}\dfrac{4}{a} + \dfrac{1}{b} + \dfrac{2}{c} = 1\\\dfrac{1}{{4a}} = \dfrac{1}{b} = \dfrac{1}{{2c}}\end{array} \right.\).

Chọn B.