Cho `x,y,z `thực thỏa `2010>=(x+1)^2+(y+2)^2+(z+3)^2` Tìm min `A=xy+y(z-1)+z(x-2)`

$\begin{array}{l} A = xy + y\left( {z - 1} \right) + z\left( {x - 2} \right)\\ A = \left( {x + 1} \right)\left( {y + 2} \right) + \left( {y + 2} \right)\left( {z + 3} \right) + \left( {z + 3} \right)\left( {x + 1} \right) - 5x - 5y - 5z - 11\\  = \left( {x + 1} \right)\left( {y + 2} \right) + \left( {y + 2} \right)\left( {z + 3} \right) + \left( {z + 3} \right)\left( {x + 1} \right) - 5\left( {x + y + z} \right) - 11\\  = \dfrac{1}{2}{\left[ {\left( {x + 1} \right) + \left( {y + 2} \right) + \left( {z + 3} \right)} \right]^2} - \dfrac{1}{2}\left[ {{{\left( {x + 1} \right)}^2} + {{\left( {y + 2} \right)}^2} + {{\left( {z + 3} \right)}^2}} \right]\\  - 5\left( {x + y + z} \right) - 11\\  = \dfrac{1}{2}{\left( {x + y + z + 6} \right)^2} - \dfrac{1}{2}\left[ {{{\left( {x + 1} \right)}^2} + {{\left( {y + 2} \right)}^2} + {{\left( {z + 3} \right)}^2}} \right]\\  - 5\left( {x + y + z} \right) - 11\\  \Rightarrow A \ge \dfrac{1}{2}{\left( {x + y + z + 6} \right)^2} - \dfrac{1}{2}.2010 - 5\left( {x + y + z} \right) - 11\\  = \dfrac{1}{2}{\left( {x + y + z + 6} \right)^2} - \dfrac{1}{2}.2010 - 5\left( {x + y + z + 6} \right) + 19\\  = \dfrac{1}{2}\left[ {{{\left( {x + y + z + 6} \right)}^2} - 10\left( {x + y + z + 6} \right) + 25} \right] - \dfrac{{25}}{2} + 19 - 1005\\  = \dfrac{1}{2}{\left( {x + y + z + 6 - 5} \right)^2} - \dfrac{{1997}}{2}\\  = \dfrac{1}{2}{\left( {x + y + z + 1} \right)^2} - \dfrac{{1997}}{2} \ge  - \dfrac{{1997}}{2}\\ ' = ' \Leftrightarrow \left\{ \begin{array}{l} x + y + z + 1 = 0\\ {\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z + 3} \right)^2} = 2010 (*)\end{array} \right. \end{array}$

(*) có thể xảy ra khi $x=-1,{z = \sqrt {3995}  - 1},x=1-\sqrt{3995}$