I need to find the points in the Cartesian plane that make $x+y+\sqrt{(x-y)^2-4}$ positive. I got a little progress but then I get stuck:
The problem is equivalent to solving $$-(x+y)<\sqrt{(x-y)^2-4}$$ If $0<-(x+y)$, then I can square both sides $$ (x+y)^2<(x-y)^2-4 \implies xy < -1 $$ Then if $x$ is positive, $y$ must be below $-\frac{1}{x}$, and if $x$ is negative then $y$ must be above it. And since $0<-(x+y) \iff y < -x$, I must also restrain $y$ to be below $-x$.
For the case $0=-(x+y)$, $y=-x$, so $x+y+\sqrt{(x-y)^2-4}=\sqrt{(2x)^2-4}$ is always positive here.
But if $-(x+y)<0$, then I can't just square the initial inequality, and I don't really know how to follow. Got any ideas? Thanks!