<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] we have $$2xy\le |x-y|\le x^2+y^2$$ now we assume $$x\geq y$$ then we get
$$2xy\le x-y$$ and $$x-y\le x^2+y^2$$
the right inequality Can be written as $$\frac{1}{2}\le (x-\frac{1}{2})^2+(y-\frac{1}{2})^2$$
the left Hand side is:
$$y(1+2x)\le x$$ 
$$x=-\frac{1}{2}$$ gives a contradiction
if $$x<-\frac{1}{2}$$ we get $$y\geq \frac{x}{1+2x}$$
if $$x>-\frac{1}{2}$$ we get $$y\le \frac{x}{1+2x}$$
Can you finish now?
</s>