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

Question

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

I could consider the eight parts the plane gets divided into by the $x$-axis, $y$-axis, $y=x$ and $y=-x$, separately and get basic inequalities free from modulus sign. Is there a faster way? Someone suggested that reverse triangle inequality could be used to conclude that $|x-y|+|x|-|y| \leq 2 \implies |y| \geq|x|-1$ but I think the graphs of these two inequalities could be different because we don't know if the latter inequality is as strong as the former. Just as $y\geq1 \implies y\geq2$ but the graphs are obviously different.

Answer 1

Just consider four cases:

1. $x\geq0,$ $y\geq0$;

2. $x\geq0$, $y\leq0$;

3. $x\leq0$, $y\geq0$ and

4. $x\leq0$, $y\leq0.$

For example, in the first case for $x-y\geq0$ we obtain $$x-1\leq y\leq x.$$

For $x\leq y$ we obtain $0\leq 2$, which says that all $y\geq x$ is valid.

Id est, we got a figure is bounded by $x=0$, $y=0$ and $y=x-1$, where $x\geq0$ and $y\geq0.$