Trong không gian với hệ tọa độ Oxyz, cho ba điểm

A

- Hướng dẫn giải

Phương trình mặt phẳng (ABC) là:

\(\begin{array}{l}\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \Rightarrow d = {d_{\left( {O,\left( P \right)} \right)}} = \frac{{\left| {\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1} \right|}}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }}\\ \Rightarrow \frac{1}{{{d^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}\\ \Rightarrow {\left( {\frac{7}{d}} \right)^2} = \left( {{a^2} + 4{b^2} + 16{c^2}} \right)\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} \right) \ge {\left( {1 + 2 + 4} \right)^2}\\ \Rightarrow \frac{7}{d} \ge 7 \Rightarrow d \le 1\end{array}\)

Dấu "=" xảy ra khi:

\(\begin{array}{l}\left( {{a^2} + 4{b^2} + 16{c^2}} \right)\left( {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} \right)\\\frac{{{a^2}}}{{\frac{1}{{{a^2}}}}} = \frac{{4{b^2}}}{{\frac{1}{{{b^2}}}}} = \frac{{16{c^2}}}{{\frac{1}{{{c^2}}}}} \Leftrightarrow {a^2} = 2{b^2} = 4{c^2}\\{a^2} + 4{b^2} + 16{c^2} = 49 \Leftrightarrow 4{c^2} + 8{c^2} + 16{c^2} = 49\\ \Rightarrow {c^2} = \frac{{49}}{{28}} = \frac{7}{4} \Rightarrow {a^2} + {b^2} + {c^2} = 7{c^2} = \frac{{49}}{4}\end{array}\)