Inequality with absolute value and variable on both sides

Question

I'm not able to understand how to get through this

$$\dfrac{|x+6|}{x+1} \leq x-2$$

$x$ can't, of course, be equal to $-1$.

I'm at a point where I'm not sure how to deal with the absolute value after reaching

$$|x+6| \leq (x-2)(x+1)$$

Thanks!

Answer 1

Lets look at this case by case.

Case 1: $x\leq-6$

Then you had $x+1\leq-5$, and so when multiplying across, the inequality sign changes. Also, $|x+6|=6-x$, so: $$6-x\geq(x-2)(x+1)=x^2-x-2\implies x^2\leq8$$ This is not possible since $x\leq -6$.

Case 2: $-6\leq x<-1$

Then you still had $x+1<0$, but now $|x+6|=x+6$, so $$x+6\geq x^2-x-2\implies x^2-2x-8\leq0\implies-2\leq x\leq 4$$ From this range, only $-2\leq x<-1$ works.

Case 3: $x>-1$

Then $$x+6\leq x^2-x-2\implies x^2-2x-8\geq0\implies x\leq-2\text{ or }x\geq4$$ From this range, only $x\geq4$ works.

---

Final solution: $x\in[-2,-1)$ or $x\in[4,\infty)$.