Sketch the region in the plane consisting of all points (x,y) such that $2xy\le |x-y|\le x^2+y^2$

Question

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?

Answer 1

$$2xy\le |x-y|\le x^2+y^2$$

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.

Answer 2

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?