<s>[INST] 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??
 [/INST] Here is a way of doing it that requires no previous knowledge of 3D analytic geometry or calculus:
You are trying to show $z = 2x^2+y^2+2xy-2x+2y+2 \gt -3$, or, (more usefully) that $z+3 = 2x^2+y^2+2xy-2x+2y+5 \gt 0$
$$z+3 = 2x^2+y^2+2xy-2x+2y+5 \\= 2x^2+(2y-2)x+(y^2+2y+5)$$
Now we can use the quadratic formula on it as if it were just a 2nd degree polynomial in $x$.
$$x = \frac{2-2y\pm \sqrt{4(y-1)^2 - 8(y^2-2y+5)}}{2} = 1-y\pm(y+3)i$$
This is clearly only a real value when $y=-3$, so a point of intersection will only occur if $y=-3$.  Another interpretation is that the vertex of the elliptic paraboloid is at $y=-3$.
We now substitute $y=-3$ into our original equation to find the parabola that represents the minimum value of the function WRT $y$:
$$z+3 = 2x^2+y^2+2xy-2x+2y+5 = 2x^2 - 8x + 8$$
We find that this has a double-root at $x=2$ (can be obtained by using quadratic formula again, although its much simpler this time).
So, we have determined that the vertex of this elliptic paraboloid is at $(x,y) = (2, -3)$.  We can now evaluate the function at this point: $z+3 = 2x^2+y^2+2xy-2x+2y+5 = 0$, or $z=-3$. Clearly, the original proposition, $z>-3$, is not correct.
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