Đáp án đúng: B
Giải chi tiết:Ta có:
\(\begin{array}{l}\,\,\,\,2\left[ {{{\left( {a + b + c} \right)}^2} - 3\left( {ab + bc + ca} \right)} \right]\\ = 2\left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca} \right]\\ = {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2}\\ \Rightarrow {\left( {a + b + c} \right)^2} - 3\left( {ab + bc + ca} \right) = \frac{1}{2}\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]\\ \Rightarrow ab + bc + ca = \frac{{{{\left( {a + b + c} \right)}^2} - \frac{1}{2}\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]}}{3}\\Do\,\,a + b + c = 1 \Rightarrow {\left( {a + b + c} \right)^2} = 1\\ \Rightarrow P = 6\left( {ab + bc + ca} \right) - 2{\left( {a + b + c} \right)^2} + a{\left( {a - b} \right)^2} + b{\left( {b - c} \right)^2} + c{\left( {c - a} \right)^2} + 2\\ \Rightarrow P = 2{\left( {a + b + c} \right)^2} - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] - 2{\left( {a + b + c} \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + a{\left( {a - b} \right)^2} + b{\left( {b - c} \right)^2} + c{\left( {c - a} \right)^2} + 2\\ \Rightarrow P = - \left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] + a{\left( {a - b} \right)^2} + b{\left( {b - c} \right)^2} + c{\left( {c - a} \right)^2} + 2\\ \Rightarrow P = {\left( {a - b} \right)^2}\left( {a - 1} \right) + {\left( {b - c} \right)^2}\left( {b - 1} \right) + {\left( {c - a} \right)^2}\left( {c - 1} \right) + 2\end{array}\)
Do \(a;b;c>0;\,a+b+c=1\Rightarrow 0
\(\Rightarrow P\le 2\)
Dấu bằng xảy ra \(\Leftrightarrow \left\{ \begin{align} & a-b=0 \\ & b-c=0 \\ & c-a=0 \\ & a+b+c=1 \\ \end{align} \right.\Leftrightarrow a=b=c=\frac{1}{3}\)
Chọn B
