show that the function $z = 2x^2 + y^2 +2xy -2x +2y +2$ is greater than $-3$ Show that the function 
$$z = 2x^2 + y^2 +2xy -2x +2y +2$$
is greater than $-3$
I tried to factorize but couldn't get more than $(x-1)^2 + (x+y)^2 +(y-1)^2 - (y)^2$.
Is there any another way to factorize or another method??

The equation $2x^2 + y^2 +2xy -2x +2y +2=0$ is a conic in the plane; so you can use the good old methods to put it in canonical form; set $x=X+a$ and $y=Y+b$, to get
$$
2X^2+4aX+2a^2+
Y^2+2bY+b^2+
2XY+2bX+2aY+2ab
-2X-2a+
2Y+2b+
2
=2X^2+2XY+Y^2+(4a+2b-2)X+(2b+2a+2)Y+2a^2+2ab+b^2-2a+2b+2
$$
and you get zero coefficients for $X$ and $Y$ by solving
$$
\begin{cases}
2a+b=1\\
a+b=-1
\end{cases}
$$
that is $a=2$ and $b=-3$. Now the polynomial becomes
$$
2X^2+2XY+Y^2-3
$$
and now it is clear that its values are $\ge-3$, because $2X^2+2XY+Y^2$ is positive definite (the conic is an ellipse), having negative discriminant $4-8=-4$. Or you can write it as
$$
X^2+(X+Y)^2-3
$$
