I need help solving this inequality with absolute values.

Question

I've been stuck on this for a while:

$$\frac{||x|-|x-2||}{x^2-2}\le2$$

I've managed to split it up into 3 parts where: $x < 0$, $0 < x < 2$ and $x > 2$.

$x<0$ and $x>2$ go smoothly but I can't get $0 < x < 2$ to work. Help.

Answer 1

Actually, first notice that $$||x|-|x-2||>0$$ for all $x\in\mathbb{R}$.

So for $-\sqrt2<x<\sqrt2$, $$x^2-2<0$$ giving the expression $$\frac{||x|-|x-2||}{x^2-2}<0\le2$$

Now this is where the brute force comes in. Consider these intervals:

• $x\le-\sqrt2$ (which is a subset of $x<0$)
• $\sqrt2\le x <2$
• $x\ge2$

You should be able to proceed from here :)

Answer 2

For $0 < x < 2$, we have $|x| = x$ and $|x - 2| = 2 - x$, hence $$\frac{||x| - |x - 2||}{x^2 - 2} = \frac{|x + 2 - x|}{x^2 - 2} = \frac{2}{x^2 - 2} \le 2$$ Equivalently, $$\frac{1}{x^2 - 2} \le 1.$$ Note that when $x^2 - 2 < 0$, i.e. where $-\sqrt{2} < x < \sqrt{2}$, then the left side is negative, and hence less than $1$. When $\sqrt{2} < x < 2$ (or when $-2 < x < -\sqrt{2}$, but don't forget, we assumed $0 < x < 2$), then the left side is positive, so we can multiply the denominator without changing sign. So, given $\sqrt{2} < x < 2$, we have $$1 \le x^2 - 2 \iff x^2 \ge 3,$$ which is true if and only if $x \ge \sqrt{3}$ or $x \le -\sqrt{3}$. Since $\sqrt{2} < x < 2$, we can discard the second possibility, so our solution when $0 < x < 2$ is $$\sqrt{3} \le x < 2.$$

Answer 3

For $x^2<2$ or $-\sqrt2<x<\sqrt2$ it's obviously true.

Let $x^2>2$.

We need to solve that $$2(x^2-2)\geq||x|-|x-2||$$ or $$4(x^4-4x^2+4)\geq x^2+(x-2)^2-2|x(x-2)|$$ or $$2x^4-9x^2+2x+6+|x(x-2)|\geq0.$$ Consider three cases:

1. $x<-\sqrt2$.

Hence, $x(x-2)>0$ and we need to sove that $$2x^4-9x^2+2x+6+x^2-2x\geq0$$ or $$x^4-4x^2+3\geq0,$$ which gives $$x\leq-\sqrt3.$$ 2. $\sqrt2<x\leq2.$

We need to sove that $$2x^4-9x^2+2x+6-x(x-2)\geq0$$ or $$x^4-5x^2+2x+3\geq0$$ or $$x^4-4x^2+4-x^2+2x-1\geq0$$ or $$(x^2-2)^2-(x-1)^2\geq0$$ or $$(x^2-x-1)(x^2+x-3)\geq0,$$ which gives $$\frac{1+\sqrt5}{2}\leq x\leq2.$$ 3. $x>2$.

We need to soove that $$2x^4-9x^2+2x+6+x(x-2)\geq0$$ or $$x^4-4x^2+3\geq0$$ or $$(x^2-1)(x^2-3)\geq0,$$ which is true for all $x>2$.

Id est, we got the answer: $$\mathbb R\setminus\left(\left(-\sqrt3,-\sqrt2\right]\cup\left[\sqrt2,\frac{1+\sqrt5}{2}\right)\right).$$