Đáp án:
$MinA = 2020 \Leftrightarrow \left( {x;y;z} \right) = \left( {2;2;1} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
A = {x^2} + {y^2} + {z^2} - yz - 4x - 3y + 2027\\
= \left( {{x^2} - 4x + 4} \right) + \left( {{z^2} - 2z.\dfrac{y}{2} + \dfrac{{{y^2}}}{4}} \right) + \dfrac{{3{y^2}}}{4} - 3y + 2023\\
= {\left( {x - 2} \right)^2} + {\left( {z - \dfrac{y}{2}} \right)^2} + \dfrac{3}{4}\left( {{y^2} - 4y + 4} \right) + 2020\\
= {\left( {x - 2} \right)^2} + {\left( {z - \dfrac{y}{2}} \right)^2} + \dfrac{3}{4}{\left( {y - 2} \right)^2} + 2020
\end{array}$
Lại có:
$\begin{array}{l}
{\left( {x - 2} \right)^2},{\left( {z - \dfrac{y}{2}} \right)^2},{\left( {y - 2} \right)^2} \ge 0,\forall x,y,z\\
\Rightarrow {\left( {x - 2} \right)^2} + {\left( {z - \dfrac{y}{2}} \right)^2} + \dfrac{3}{4}{\left( {y - 2} \right)^2} \ge 0,\forall x,y,z\\
\Rightarrow {\left( {x - 2} \right)^2} + {\left( {z - \dfrac{y}{2}} \right)^2} + \dfrac{3}{4}{\left( {y - 2} \right)^2} + 2020 \ge 2020,\forall x,y,z\\
\Rightarrow MinA = 2020
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow {\left( {x - 2} \right)^2} = {\left( {z - \dfrac{y}{2}} \right)^2} = {\left( {y - 2} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
z = \dfrac{y}{2}\\
y = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 2\\
z = 1
\end{array} \right.
\end{array}$
Vậy $MinA = 2020 \Leftrightarrow \left( {x;y;z} \right) = \left( {2;2;1} \right)$
