show that the function $z = 2x^2 + y^2 +2xy -2x +2y +2$ is greater than $-3$

Question

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??

Answer 1

HINT:

$$z = 2x^2 + y^2 +2xy -2x +2y +2=(x+y+1)^2+(x-2)^2+2-4-1$$

Derivation :

Let $$ 2x^2 + y^2 +2xy -2x +2y +2=(x+y+a)^2+(x+b)^2+2-a^2-b^2$$

$$\implies 2x^2 + y^2 +2xy -2x +2y +2=2x^2+y^2+2xy+2x(a+b)+2ay+2-a^2-b^2$$

Comparing the coefficients of $y, a=1$

Comparing the coefficients of $x, a+b=-1\implies b=-1-a=-2$

Answer 2

If you don't want to look for a factorisation, you can always use some analysis:

Let $q(x,y)=2x^2+y^2+2xy-2x+2y+2$. Then $$\frac{\partial q}{\partial x}(x,y)=4x+2y-2,\ \frac{\partial q}{\partial y}(x,y)=2y+2x+2.$$ Therefore the unique critical point of $q$ is $(2,-3)$ which has image $q(2,-3)=-3$. Now it only remains to see that this point is a minimum. Indeed, the corresponding Hessian matrix is $$ \begin{pmatrix} 4 & 2\\ 2 & 2\\ \end{pmatrix}, $$ which is positive defined.

Answer 3

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 $$ X^2+(X+Y)^2-3 $$

Answer 4

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 2^nd 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.

Answer 5

$$z=2x^2 + y^2 +2xy -2x +2y +2$$

$$\implies 2x^2+2x(y-1)+y^2+2y+2-z=0$$

As $x$ is real, the discriminant of the above Quadratic equation must be $\ge0$

$$\implies \{2(y-1)\}^2-4\cdot 2\cdot(y^2+2y+2-z)\ge0 $$

$$\implies (y-1)^2- 2\cdot(y^2+2y+2-z)\ge0 $$

$$\implies 2z\ge y^2+6y+3$$

$$\text{Now, }y^2+6y+3=(y+3)^2-6\ge -6$$