Đáp án đúng: C

Phương pháp giải:

Vì \(a,\,\,b,\,\,c \in \left[ { - 1;\,\,1} \right]\) nên \(0 \le {a^2},\,\,{b^2},\,\,{c^2} \le 1\).
Từ bất đẳng thức \(\left( {1 - {b^2}} \right)\left( {1 + {b^2} - {a^4}} \right) \ge 0\) chứng minh được \({a^4} + {b^4} - {a^4}{b^2} \le 1\). Từ đó suy ra, \({a^{2020}} + {b^{2020}} \le 1 + {a^4}{b^2}\).
Chứng minh tương tự đối với \({b^{2020}} + {c^{2020}} \le 1 + {b^4}{c^2},\,\,{c^{2020}} + {a^{2020}} \le 1 + {c^4}{a^2}\).Giải chi tiết:Vì \(a,\,\,b,\,\,c \in \left[ { - 1;\,\,1} \right]\) nên \(0 \le {a^2},\,\,{b^2},\,\,{c^2} \le 1\).
\( \Leftrightarrow \left\{ \begin{array}{l}1 + {b^2} - {a^4} - {b^2} - {b^4} + {a^4}{b^2} \ge 0\\1 + {c^2} - {b^4} - {c^2} - {c^4} + {c^2}{b^4} \ge 0\\1 + {a^2} - {c^4} - {a^2} - {a^4} + {a^2}{c^4} \ge 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{a^4} + {b^4} - {a^4}{b^2} \le 1\\{b^4} + {c^4} - {b^4}{c^2} \le 1\\{c^4} + {a^4} - {c^4}{a^2} \le 1\end{array} \right.\)\( \Rightarrow \left\{ \begin{array}{l}\left( {1 - {b^2}} \right)\left( {1 + {b^2} - {a^4}} \right) \ge 0\\\left( {1 - {c^2}} \right)\left( {1 + {c^2} - {b^4}} \right) \ge 0\\\left( {1 - {a^2}} \right)\left( {1 + {a^2} - {c^4}} \right) \ge 0\end{array} \right.\) Mà ta lại có: \(\left\{ \begin{array}{l}{a^4} \ge {a^{2020}}\\{b^4} \ge {b^{2020}}\\{c^4} \ge {c^{2020}}\end{array} \right.\) đúng với mọi \(a,\,\,b,\,\,c \in \left[ { - 1;\,\,1} \right]\).
\( \Rightarrow \left\{ \begin{array}{l}{a^{2020}} + {b^{2020}} - {a^4}{b^2} \le {a^4} + {b^4} - {a^4}{b^2} \le 1\\{b^{2020}} + {c^{2020}} - {b^4}{c^2} \le {b^4} + {c^4} - {b^4}{c^2} \le 1\\{c^{2020}} + {a^{2020}} - {c^4}{a^2} \le {c^4} + {a^4} - {c^4}{a^2} \le 1\end{array} \right.\)
\( \Rightarrow \left\{ \begin{array}{l}{a^{2020}} + {b^{2020}} \le 1 + {a^4}{b^2}\\{b^{2020}} + {c^{2020}} \le 1 + {b^4}{c^2}\\{c^{2020}} + {a^{2020}} \le 1 + {c^4}{a^2}\end{array} \right.\)
\( \Rightarrow \left\{ \begin{array}{l}{a^{2020}} + {b^{2020}} + {c^{2020}} \le 1 + {a^4}{b^2} + {c^{2020}}\\{a^{2020}} + {b^{2020}} + {c^{2020}} \le 1 + {b^4}{c^2} + {a^{2020}}\\{c^{2020}} + {a^{2020}} + {b^{2020}} \le 1 + {c^4}{a^2} + {b^{2020}}\end{array} \right.\)
\( \Rightarrow 3\left( {{a^{2020}} + {b^{2020}} + {c^{2020}}} \right) \le 3 + \left( {{a^4}{b^2} + {b^4}{c^2} + {c^4}{a^2}} \right) + \left( {{a^{2020}} + {b^{2020}} + {c^{2020}}} \right)\)
\( \Rightarrow 2\left( {{a^{2020}} + {b^{2020}} + {c^{2020}}} \right) \le 3 + \left( {{a^4}{b^2} + {b^4}{c^2} + {c^4}{a^2}} \right)\)
\( \Rightarrow {a^{2020}} + {b^{2020}} + {c^{2020}} \le \dfrac{{3 + {a^4}{b^2} + {b^4}{c^2} + {c^4}{a^2}}}{2}\)
\(\begin{array}{l} \Rightarrow {a^{2020}} + {b^{2020}} + {c^{2020}} \le \dfrac{{3 + {a^4}{b^2} + {b^4}{c^2} + {c^4}{a^2}}}{2}\\ \Rightarrow 2 \le \dfrac{{3 + {a^4}{b^2} + {b^4}{c^2} + {c^4}{a^2}}}{{{a^{2020}} + {b^{2020}} + {c^{2020}}}}\end{array}\)
\( \Rightarrow 2 \le \dfrac{{3 + {a^4}{b^2} + {b^4}{c^2} + {c^4}{a^2}}}{{{a^{2020}} + {b^{2020}} + {c^{2020}}}}\)
Vậy \(\dfrac{{3 + {a^4}{b^2} + {b^4}{c^2} + {c^4}{a^2}}}{{{a^{2020}} + {b^{2020}} + {c^{2020}}}} \ge 2\) với \(a,\,\,b,\,\,c\) thuộc \(\left[ { - 1;\,\,1} \right]\).
Chọn C.