Đáp án:
$\begin{array}{l}
12a + {\left( {b - c} \right)^2}\\
= 4a.3 + {\left( {b - c} \right)^2}\\
= 4a\left( {a + b + c} \right) + {\left( {b - c} \right)^2}\\
= 4{a^2} + 4ab + 4ac + {b^2} - 2bc + {c^2}\\
Giả\,sử:a \le b \le c\\
\Rightarrow a \le b \le c \le 1\\
Áp\,dụng:{\left( {x + y + z} \right)^2} \le 3\left( {{x^2} + {y^2} + {z^2}} \right)\\
\Rightarrow {K^2} = {\left( {\sqrt {12a + {{\left( {b - c} \right)}^2}} + \sqrt {12b + {{\left( {a - c} \right)}^2}} + \sqrt {12c + {{\left( {a - b} \right)}^2}} } \right)^2}\\
\Rightarrow {K^2} \le 3\left( {12a + {{\left( {b - c} \right)}^2} + 12b + {{\left( {a - c} \right)}^2} + 12c + {{\left( {a - b} \right)}^2}} \right)\\
\Rightarrow {K^2} \le 3\left( \begin{array}{l}
4{a^2} + 4ab + 4ac + {b^2} - 2bc + {c^2} + \\
4{b^2} + 4ab + 4bc + {c^2} - 2ac + {a^2} + \\
4{c^2} + 4bc + 4ac + {a^2} - 2ab + {b^2}
\end{array} \right)\\
\Rightarrow {K^2} \le 18\left( {{a^2} + {b^2} + {c^2} + ab + bc + ac} \right)\\
\Rightarrow {K^2} \le 9\left( {{a^2} + {b^2} + {c^2}} \right) + 9{\left( {a + b + c} \right)^2}\\
\Rightarrow {K^2} \le 9.3 + {9.3^2}\\
\Rightarrow K \le 6\sqrt 3 \\
Dấu\, = \,xảy\,ra \Leftrightarrow a = b = c = 1
\end{array}$
Vậy GTLN của bt là $6\sqrt 3 $
