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 $xp<0$?
 A: 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 $x+p\neq 0$.
A: A possible way by splitting in only two branches is as follows using the inequality between arithmetic and geometric mean:


*

*AM-GM: $a,b\geq 0 \Rightarrow \frac{a+b}{2}\geq \sqrt{ab}$ and equality holds if and only if $a= b$ (which can be  easily proved by squaring and rearranging).


Cancelling gives


*

*$\frac{x^{2}+p^{2}}{xp} = \frac{x}{p} + \frac{p}{x}$

*Setting $y:=\frac{x}{p}$, you get for $xp>0 \Leftrightarrow y= \frac{x}{p}>0$ the inequality 
$$y+\frac{1}{y}\stackrel{AM-GM}{>} 2 \Leftrightarrow y\neq 1,y>0$$
It follows immediately for $px<0\Leftrightarrow y= \frac{x}{p}<0$
$$y+\frac{1}{y} < -2 \Leftrightarrow y\neq -1,y<0 \Leftrightarrow  \frac{x}{p}\neq-1,\frac{x}{p}<0 $$
So, the solutions are


*

*$p>0$: $x\in (-\infty,0)\setminus\{-p\}$

*$p<0$: $x\in (0,+\infty)\setminus\{-p\}$
A: Don't overcomplicate things.  You correctly infer that you would like to cancel the $xp$ term, and that this cancelation might change the direction of the inequality, depending on the sign of $xp$.  However, breaking things into four cases is too much work.  Instead, why not simply consider two cases:  (1) $xp > 0$, and (2) $xp < 0$ (note that we cannot have $xp = 0$, as the original rational expression would have been undefined in the first place).  We might have to break these cases up more later, depending on where the computation leads us, but why make life messy earlier than we have to?


*

*If we assume that $xp > 0$, then we obtain
$$ x^2 + p^2 < -2xp < 0, $$
where the final inequality follows from the fact that $xp$ is positive.  Since $x^2$ and $p^2$ are both necessarily nonnegative, this is impossible.  This implies that $xp$ cannot be nonnegative.  That is, this first case cannot occur, and we don't have to do any more work here.

*If, on the other hand, we assume that $xp < 0$, then we obtain
$$ x^2 + p^2 > -2xp
\implies x^2 + 2xp + p^2 > 0
\implies (x+p)^2 > 0. $$
This is true for all $x \ne -p$.  Therefore $xp$ must be negative and we must have $x \ne -p$.
Combining the above results, we get the set of solutions
$$ \{ x\in \mathbb{R} \mid xp < 0 \text{ and } x \ne -p \}. $$
Alternatively, we could describe this as the set of $x$ such that the sign of $x$ differs from the sign of $p$, and $x\ne -p$.
