Is the set $M=\{[x,y,z]\in{\mathbf R}^3 :\ x^2 + y^2 +z^2 + xy + yz + xz = 1,\ x \ge 0,\ y \ge 0\}$ compact? How do I show that the set
$$
M=\{[x,y,z]\in{\mathbf R}^3 :\ x^2 + y^2 +z^2 + xy + yz + xz = 1,\ x \ge 0,\ y \ge 0\}
$$
is compact?
The set is apparently closed (as it cannot be open given that the condition defining $M$ is a continuous function).
The exercise is to compute extrema of
$$
f(x,y,z) = \sqrt{x^2+y^2+z^2}
$$
over this set. Simply assuming it is compact, I suspect maximum is attained for points in ${\mathbf R^3}$ satisfying
$$
x^2 + xy + y^2 = 1\quad {\rm and}\quad z = -(x+y).
$$
That's what I need. The calculation wasn't hard using Lagrange's multipliers method assuming $M$ was compact, but I am unsure as how to proceed in showing it actually is.
Any help will be appreciated.
I noticed that for $z \ge 0$, this is simple as all the terms in the first condition defining $M$ are greater than or equal to zero, thus it holds that
$$
x^2 + y^2 +z^2 + xy + yz + xz \ge x^2 + y^2 + z^2
$$
and so $M_{z \ge 0} \subset B([0,0,0], 1)$. But for the $z < 0$ case I just don't see why the set is bounded; surely it is just as easy as the former case.
 A: From
$$f(x,y,z):=x^2+y^2+z^2+xy+yz+zx={1\over2}\bigl((x+y+z)^2+x^2+y^2+z^2\bigr)$$
it follows that
$$x^2+y^2+z^2\leq 2f(x,y,z)=2\qquad\forall\ (x,y,z)\in M\ .$$
This shows that $M$ is (not only closed, but) also bounded, hence compact.
A: Note that $x^2 + y^2 + z^2 + xy + xz + yz = {\bf x}^TA{\bf x}$ where ${\bf x} = [x\ \ y\ \ z]^T$ and
$$A = \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & 1 & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} & 1\end{bmatrix}.$$
As $A$ is symmetric, we can orthogonally diagonalise $A$. That is, we can find an orthogonal matrix $P$ such that $P^TAP = \operatorname{diag}(\lambda_1, \lambda_2, \lambda_3)$ where the $\lambda_i$ are the eigenvalues of $A$. Making the change of variables ${\bf x} = P{\bf\hat{x}}$, the equation ${\bf x}^TA{\bf x} = 1$ becomes ${\bf \hat{x}}^T\operatorname{diag}(\lambda_1, \lambda_2, \lambda_3){\bf\hat{x}} = 1$. Writing ${\bf\hat{x}} = [\hat{x}\ \ \hat{y}\ \ \hat{z}]$, this becomes $\lambda_1\hat{x}^2 + \lambda_2\hat{y}^2 + \lambda_3\hat{z}^2 = 1$. This is the standard way to determine the type of a  quadric.
In this case, $\lambda_1 = 2$, $\lambda_2 = \lambda_3 = \frac{1}{2}$ (the ordering of eigenvalues is not particularly important), so the equation becomes $2\hat{x}^2 + \frac{1}{2}\hat{y}^2 + \frac{1}{2}\hat{z}^2 = 1$. The set of points $(\hat{x}, \hat{y}, \hat{z})$ which satisfy this equation is an ellipsoid, which is compact.
So $E = \{(x, y, z) \mid x^2 + y^2 + z^2 + xy + yz+ xz = 1\} = \{(\hat{x}, \hat{y}, \hat{z}) \mid 2\hat{x}^2 + \frac{1}{2}\hat{y}^2 + \frac{1}{2}\hat{z}^2 = 1\}$ is compact, in particular, closed and bounded. As $C = \{(x, y, z) \mid x \geq 0, y \geq 0\}$ is closed, $M = E\cap C$ is closed, and as $M \subset E$ and $E$ is bounded, $M$ is bounded. Therefore, $M$ is compact.
A: Let us define $h(x,y,z) = x^2 + y^2 + z^2 + xy + yz + xz$. The condition in question for $M$ is that for $[x,y,z] \in M$ it holds that $h(x,y,z) = 1$.
For $z \ge 0$ we know that the set $M_{z \ge 0}$ fits into a unit ball. The part of the set for $z \le 0$ is a point-reflection of $M_{z \ge 0}$, because $h(x, y, z) = h(-x, -y, -z)$. So the other part fits into a unit ball as well.
(For any point with $z$ negative, we may find a point with $z$ positive and we have shown all such points are located in a unit ball.)
