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

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.

• You can take this fact in use that if $(x,y)$ belongs to the plot, then so does $(-x,-y)$ and then reduce the number of regions by twice. – Mostafa Ayaz Mar 30 at 8:30
• The $|x| \leq 1$ satisfies the inequality, since $|x-y|+|x|-|y| \leq |x|+|y|+|x|-|y| \leq 2$, so at least that leaves you with $|x|>1$. Combine it with Mostafa's comment and you "only" need to check $x>1$. (I still think your solution is okay though) – Sil Mar 30 at 8:31

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 $$y\geq0.$$