Help solving inequality in two variables involving a square root

Question

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!

Answer 1

You need $|x-y| \ge 2$ to make the square root real. If $|x-y| \ge 2$ and $x+y > 0$, the inequality is satisfied. If $|x-y| \ge 2$ and $x-y \le 0$, you need $xy < -1$ as you said.

So above/right of the line $x+y=0$ we have the regions $y \ge x+2$ and $y \le x-2$. Below/left of the line $x+y=0$ we have $y > -1/x$ for $x < 0$ and $y < -1/x$ for $x > 0$. It looks something like the blue-shaded regions here, where $x+y=0$ is the dotted line.

Answer 2

As pointed out by Josb B. in the comments, indeed if ${-(x+y) < 0}$ then because the square root is always going to be positive (the square root always spits out the principle root) - then the inequality automatically holds so long as the argument inside the root is positive. Otherwise the expression becomes undefined in the context of real numbers. This is I think the last constraint you need to ensure. The intersection of all such constraints should give you a region in ${\mathbb{R}^2}$ for which the inequality holds