# Help solving inequality in two variables involving a square root

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!

• If $-(x+y)<0$, then the inequality is always valid so long as the right side has a square root, since the square root is necessarily positive. Thus, the condition of $-(x+y)<0$ boils down to $(x-y)^2-4\geq 0$. Jun 24, 2020 at 20:12

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.
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