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??

$$z=2x^2 + y^2 +2xy -2x +2y +2$$
$$\implies 2x^2+2x(y-1)+y^2+2y+2-z=0$$
As $x$ is real, the discriminant of the above Quadratic equation must be $\ge0$
$$\implies \{2(y-1)\}^2-4\cdot 2\cdot(y^2+2y+2-z)\ge0 $$
$$\implies (y-1)^2-  2\cdot(y^2+2y+2-z)\ge0 $$
$$\implies 2z\ge y^2+6y+3$$
$$\text{Now, }y^2+6y+3=(y+3)^2-6\ge -6$$
