Find $x \in \mathbb{R}$ such that $(x^2+p^2)/xp < -2$ I'm trying to solve the following inequality for $x\in\mathbb{R}$: $$\frac{x^{2}+p^{2}}{xp} < -2, $$
where $p$ is a real parameter.  I want to have my $x$ on the left side and everything else on the right side.
The first step I would take is to multiply both sides of the inequality with $xp$, however I'm not sure how to do so. Do I need to split my solution into 4 branches (1. where $x>0$ and $p<0$, 2. where $x<0$ and $p>0$, 3. where $x>0$ and $p>0$ and 4. where $x<0$ and $p<0$), or can I just simple split it into two branches, where $xp>0$ and where $xp<0$?

I suggest to proceed as follows
$$\frac{x^{2}+p^{2}}{xp} < -2\iff \frac{x^{2}+p^{2}}{xp} +2<0\iff \frac{x^{2}+2xp+p^{2}}{xp}<0\iff \frac{(x+p)^2}{xp}<0$$
and the sign corresponds to the sign of $xp\ne 0$.
Proceeding by multiplication we have


*

*$xp>0$
$$\frac{x^{2}+p^{2}}{xp} < -2\iff x^2+p^2<-2px \iff (x+p)^2<0$$
which is not possible, and 


*

*$xp<0$
$$\frac{x^{2}+p^{2}}{xp} < -2\iff x^2+p^2>-2px \iff (x+p)^2>0$$
which is always true, then $xp<0$ is the solution.
As noticed by Xander Henderson, of course the given solution holds for $x+p\neq 0$.
