Prove:$x^2+2xy+3y^2-6x-2y\ge-11\;\;\forall(x,y)\in\Bbb{R}$

Question

Problem Statement:-

Prove that for all real values of $x$ and $y$ $$x^2+2xy+3y^2-6x-2y\ge-11$$

I have no idea how to approach this question all I could think on seeing it was tryin to find the linear factors, turns out that the determinant $$\begin{vmatrix} a & h & g\\ h & b & f\\ g & f & c\\ \end{vmatrix}\neq0$$

So finding linear factor just flew out of the window, so I plotted the equation $x^2+2xy+3y^2-6x-2y=0$ turns out it is an ellipse. I could conclude no further as to how to approach this problem.

Answer 1

We have $$\begin{align}x^2+2xy+3y^2-6x-2y+11&=x^2+(2y-6)x+3y^2-2y+11\\\\&=(x+y-3)^2-(y-3)^2+3y^2-2y+11\\\\&=(x+y-3)^2+2y^2+4y+2\\\\&=(x+y-3)^2+2(y+1)^2\end{align}$$

Answer 2

Consider $f(x,y) = x^2 + 2xy + 3y^2 -6x -2y + 11 = x^2 + (2y -6)x + (3y^2-2y -11)$ as quadratic polynomial on $x$ variable, then the classic discriminant is $$ \Delta =-8y^2 -16y -8 < 0 \quad\forall y\in \mathbb{R}/\{-1\} $$ Thus $f(x,y) > 0 $ for each $y \in \mathbb{R}/\{ -1\}$, now the only case that you should work is for $y=-1$ and this is easy.

Answer 3

Once you can use derivatives this is straightforward:

$$ f_x = 2x+2y-6=0\\ f_y = 2x+6y-2=0 $$

These give you a critical point $(x_0,y_0)=(4,-1)$. You can verify that $f(4,-1)=-11$. It remains to show that this is indeed the minimum (here is where the second derivatives come):

$$ f_{xx} = 2,f_{xy} = 2,f_{yy} = 6 $$

So you have

$$ f_{xx}f_{yy} -f_{xy}^2 = 12-4=8 > 0, f_{xx}>0 $$

and by the Second Derivative Test it is sufficient to conclude that $f(4,-1)=-11$ is indeed the minimum.

Answer 4

Let $f(x,y)=x^2+2xy+3y^2-6x-2y$

$$\frac{\partial f(x,y)}{\partial x}=2x+2y-6=0\\\frac{\partial f(x,y)}{\partial y}=6y+2x-2=0$$ It is clear that $f(x,y)$ has no maximum. Then the solution of this system $(x,y)=(4,-1)$ gives a minimum. This minimum is precisely $$f(4,-1)=-11$$

Answer 5

The matrix for the quadratic form is $$ A=\begin{bmatrix} 1 & 1 & -3 \\ 1 & 3 & -1 \\ -3 & -1 & 11 \end{bmatrix} $$ A sufficient condition for the quadratic form to be positive semidefinite is that the leading principal minors but the last (that ought to be nonnegative) are positive. If also the full determinant is positive, the matrix is positive definite.

The leading principal minors are $$ \det[1]=1, \qquad \det\begin{bmatrix}1 & 1 \\ 1 & 3\end{bmatrix}=2, \qquad \det A=0 $$ Thus $A$ is a positive semidefinite symmetric matrix and thus $v^TAv\ge0$ for all $v\in\mathbb{R}^3$, in particular for $v=[x\;y\;1]^T$, which proves the thesis.

The null space can be determined by Gaussian elimination: \begin{align} \begin{bmatrix} 1 & 1 & -3 \\ 1 & 3 & -1 \\ -3 & -1 & 11 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & -3 \\ 0 & 2 & 2 \\ 0 & 2 & 2 \end{bmatrix}&&R_2\gets R_2-R_1, R_3\gets R_3+3R_1 \\&\to \begin{bmatrix} 1 & 1 & -3 \\ 0 & 1 & 1 \\ 0 & 2 & 2 \end{bmatrix}&&R_2\gets \tfrac{1}{2}R_2 \\&\to \begin{bmatrix} 1 & 1 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}&&R_3\gets R_3-2R_2 \\&\to \begin{bmatrix} 1 & 0 & -4 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}&&R_1\gets R_1-R_2 \end{align} So an eigenvector for the zero eigenvalue is $$ \begin{bmatrix}4 \\ -1 \\ 1\end{bmatrix} $$ and, indeed, the point with coordinates $(4,-1)$ is the only one where the form vanishes.