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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.

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    $\begingroup$ 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. $\endgroup$ – Mostafa Ayaz Mar 30 at 8:30
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    $\begingroup$ 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) $\endgroup$ – Sil Mar 30 at 8:31
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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 $y\geq0.$

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