<s>[INST] Sketch the region in the plane consisting of all points (x,y) such that $2xy\le |x-y|\le x^2+y^2$ Question: Sketch the region in the plane consisting of all points (x,y) such that $2xy\le |x-y|\le x^2+y^2$
My attempt:
If $x>y$ then
$|x-y|=x-y$
Thus now,
$2xy\le x-y\le x^2+y^2$
If $x-y\geq 2xy$ then let $x-y=k$ ($k$ is some real number)
then $y=x-k \tag{1}$ 
Thus the straight line (1) has a negative slope. 
After this how to proceed?
 [/INST] $$2xy\le |x-y|\le x^2+y^2$$
First, consider the inequality on the right.
Depending on whether $x$ or $y$ is the larger of the two the second inequality resolves into the two inequalities
$$ \left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2\ge\left(\frac{\sqrt{2}}{2}\right)^2 $$
or
$$ \left(x+\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2\ge\left(\frac{\sqrt{2}}{2}\right)^2 $$
So the solution set must lie outside the interiors of these two circles.
The first inequality resolves into the two inequalities
$$ 2xy\le x-y \text{ given that } x\ge y$$
$$ 2xy\le y-x\text{ given that } x\le y $$
The first inequality can be re-written as
\begin{eqnarray}
2xy-x+y &\le& 0\\
xy-\frac{1}{2}x+\frac{1}{2}y-\frac{1}{4} &\le& -\frac{1}{4}\\
\left(x+\frac{1}{2}\right)\left(y-\frac{1}{2}\right) &\le& -\frac{1}{4} \text{ and }y\le x
\end{eqnarray}
The graph of this inequality consists of all points lying on or below the right branch of the hyperbola.
Similiarly, the second inequality can be re-formed as
$$ \left(x-\frac{1}{2}\right)\left(y+\frac{1}{2}\right) \le -\frac{1}{4} \text{ and }y\ge x$$
The graph of this inequality consists of all points lying on or above the left branch of this hyperbola.
The darker region in the following is the solution set.

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