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.