Tìm GTNN : `A = (x-y)^2 + (x-1)^2 + (y+1)^2 + 12`

$A = (x-y)^2 + (x-1)^2 + (y+1)^2+ 12$

$\Leftrightarrow A = 2x^2 - 2xy + 2y^2 - 2x + 2y + 14$

$\Leftrightarrow 2A = 4x^2 - 4xy + 4y^2 - 4x + 4y + 28$

$\Leftrightarrow 2A = (4x^2 - 4xy + y^2 - 4x + 2y + 1) + 3\left(y^2 + 2y\cdot \dfrac13 + \dfrac19\right) + \dfrac{80}{3}$

$\Leftrightarrow 2A = (-2x + y + 1)^2 + 3\left(y + \dfrac13\right)^2 + \dfrac{80}{3}$

$\Leftrightarrow A = \dfrac12(-2x + y + 1)^2 + \dfrac32\left(y + \dfrac13\right)^2 + \dfrac{40}{3}$

Ta có:

$\begin{cases}(-2x + y + 1)^2\geqslant 0,\ \forall x;y\\\left(y + \dfrac13\right)^2 \geqslant 0,\ \forall y\end{cases}$

Do đó:

$\dfrac12(-2x + y + 1)^2 + \dfrac32\left(y + \dfrac13\right)^2 + \dfrac{40}{3} \geqslant \dfrac{40}{3}$

Hay $A\geqslant \dfrac{40}{3}$

Dấu $=$ xảy ra $\Leftrightarrow \begin{cases}-2x + y + 1 =0\\y + \dfrac13 = 0\end{cases}\Leftrightarrow \begin{cases}x = \dfrac13\\y = - \dfrac13\end{cases}$

Vậy GTNN của $A$ là $\dfrac{40}{3}$ khi $(x,y) = \left(\dfrac13,-\dfrac13\right)$