Đáp án:
$\begin{array}{l}
12a + {\left( {b - c} \right)^2} = 4a\left( {a + b + c} \right) + {\left( {b - c} \right)^2}\\
= 4{a^2} + 4ab + 4ac + {b^2} + {c^2} - 2bc\\
Giả\,sử:a \le b \le c \le 1\\
AD:{\left( {x + y + z} \right)^2} \le 3\left( {{x^2} + {y^2} + {z^2}} \right)\\
A = \sqrt {12a + {{\left( {b - c} \right)}^2}} + \sqrt {12b + {{\left( {c - a} \right)}^2}} + \sqrt {12c + {{\left( {a - b} \right)}^2}} \\
\Rightarrow {A^2} \le 3\left( \begin{array}{l}
4{a^2} + 4ab + 4ac + {b^2} + {c^2} - 2bc + \\
4{b^2} + 4bc + 4ba + {c^2} + {a^2} - 2ac + \\
4{c^2} + 4ca + 4bc + {a^2} + {b^2} - 2ab
\end{array} \right)\\
\Rightarrow {A^2} \le 18\left( {{a^2} + {b^2} + {c^2} + ab + bc + ca} \right)\\
\Rightarrow {A^2} \le 9\left( {{a^2} + {b^2} + {c^2}} \right) + 9{\left( {a + b + c} \right)^2}\\
\Rightarrow {A^2} \le 9.3 + {9.3^2} = 108\\
\Rightarrow A \le 6\sqrt 3 \\
\Rightarrow GTLN:A = 6\sqrt 3 \Leftrightarrow a = b = c = 1
\end{array}$
