Lagrange Method Optimization Problem with constraint The problem is as follows: 
$$\min_{x,y} x + y$$
subject to 
$$xy = 0.25$$
My attempt: 
I used the method of Lagrange multipliers here setting
$f(x,y) = x + y$ and $g(x,y) = xy - 0.25 = 0$
So we have $\nabla f = \lambda \nabla f$ 
$\implies 1 = \lambda y$, and $1 = \lambda x$
$\implies x = y$ and from the constraint we get:
$x^2 = 0.25 \implies x = \pm \sqrt{0.25} = y$
Clearly we choose $x = y = - \sqrt{0.25}$ to minimize the function.
However, I checked this minimization problem with Wolfram Alpha and it says no global minima where found. Why is this and where did I go wrong? Thanks!
 A: The points where $xy=\frac14$ are shown in the blue hyperbolas below.

The red and yellow lines are the points where $x+y=1$ and $x+y=-1$.
The line $x+y=1$ is the furthest to the lower left that intersects the branch of the hyperbola in quadrant $1$. So if we restrict our attention to $x,y\gt0$, we get $x+y\ge1$. Thus, there is a minimum.
If we also consider $x,y\lt0$, then the line $x+y=-1$ is the furthest to the upper right that intersects the branch of the hyperbola in quadrant $3$. For the points in quadrant $3$, we have $x+y\le-1$. In this case, there is no minimum.

Note that if $x,y\gt0$, then
$$
x+y-2\sqrt{xy}=\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\tag{1}
$$
Since $xy=\frac14$, $(1)$ implies
$$
x+y\ge1\tag{2}
$$
A: As pointed out in the Comments under the question, if $x,y$ are allowed to be negative, we can get $x+y$ as small as we wish ($x=-N,y=-1/4N$). If we require $x,y$ to be positive, then AM/GM gives $\frac{x+y}{2}\ge\sqrt{xy}=\frac{1}{2}$, so $x+y\ge 1$ with equality when $x=y=\frac{1}{2}$.
