Find $x \in \mathbb{R}$ such that $(x^2+p^2)/xp < -2$

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 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

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