2
$\begingroup$

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!

$\endgroup$
1
  • 1
    $\begingroup$ 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 $(x-y)^2-4\geq 0$. $\endgroup$
    – Josh B.
    Jun 24, 2020 at 20:12

2 Answers 2

2
$\begingroup$

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.

enter image description here

$\endgroup$
1
  • $\begingroup$ +1 for the figure. $\endgroup$ Jun 24, 2020 at 22:15
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.