Minimize the expression!

One more method of doing it except completing the square.

First differentiate w.r.t to $x$ and set it equal to $0$ .Repeat the same w.r.t $y$ .

$\large d/dx$ $\large(x^2 + 2xy + 3y^2 + 2x + 6y + 4)$ = $\large 2x+2y + 2$ $= 0$ $...(1)$

$\large d/dy$ $\large(x^2 + 2xy + 3y^2 + 2x + 6y + 4)$ = $\large 2x+6y + 6$ $= 0$ $...(2)$

Solving 1 and 2,

$\Huge \color{#D61F06}{x_{min}=0}$

$\Huge \color{#3D99F6}{y_{min}=-1}$

$\Huge \color{#20A900}{z_{min}=1}$

$z=(x+y+1)^2 +2(y+1)^2 + 1$

$\text{Clearly at }$ $\color{#D61F06}{x=0}$ and $\color{#D61F06}{y= -1} ,\text{both square expression becomes zero , hence :}$ $\LARGE \color{#EC7300}{\boxed{z_{min}=1}}$