Đáp án:

\[{G_{\min }} =  - 49 \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = \dfrac{1}{4}\\
a = 40
\end{array} \right.\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
G = {\left( {x - ay} \right)^2} + 6.\left( {x - ay} \right) + {x^2} + 16{y^2} - 8ay + 2x - 8y + 10\\
 = {\left( {x - ay} \right)^2} + 6x - 6ay + {x^2} + 16{y^2} - 8ay + 2x - 8y + 10\\
 = {\left( {x - ay} \right)^2} + 8x - 14ay + {x^2} + 16{y^2} - 8y + 10\\
 = {\left( {x - ay} \right)^2} + \left( {14x - 14ay} \right) + {x^2} + 16{y^2} - 6x - 8y + 10\\
 = \left[ {{{\left( {x - ay} \right)}^2} + 14.\left( {x - ay} \right) + 49} \right] + \left( {{x^2} - 6x + 9} \right) + \left( {16{y^2} - 8y + 1} \right) - 49\\
 = \left[ {{{\left( {x - ay} \right)}^2} + 2.\left( {x - ay} \right).7 + {7^2}} \right] + \left( {{x^2} - 2.x.3 + {3^2}} \right) + \left[ {{{\left( {4y} \right)}^2} - 2.4y.1 + {1^2}} \right] - 49\\
 = {\left( {x - ay + 7} \right)^2} + {\left( {x - 3} \right)^2} + {\left( {4y - 1} \right)^2} - 49\\
{\left( {x - ay + 7} \right)^2} \ge 0,\,\,\,\,\forall a,x,y\\
{\left( {x - 3} \right)^2} \ge 0,\,\,\,\,\forall x\\
{\left( {4y - 1} \right)^2} \ge 0,\,\,\,\,\forall y\\
 \Rightarrow {\left( {x - ay + 7} \right)^2} + {\left( {x - 3} \right)^2} + {\left( {4y - 1} \right)^2} - 49 \ge  - 49\\
 \Rightarrow G \ge  - 49,\,\,\,\,\forall a,x,y\\
 \Rightarrow {G_{\min }} =  - 49 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - ay + 7} \right)^2} = 0\\
{\left( {x - 3} \right)^2} = 0\\
{\left( {4y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = \dfrac{1}{4}\\
a = 40
\end{array} \right.
\end{array}\)

Vậy \({G_{\min }} =  - 49 \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = \dfrac{1}{4}\\
a = 40
\end{array} \right.\)