# How to solve $|x^2-1|-2\ge 2x$

I am trying to solve the inequality

$$|x^2-1|-2\ge 2x$$

but I am not sure where to start, because I have $x²$ in the absolute value part. $x²$ is always positive, and this is confusing for me. Can you please explain how to solve it?

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Consider two cases, $x^2>1$ and $x^2\le1$. – Macavity Feb 10 '14 at 14:27
$x^2$ is always positive, but $x^2 - 1$ (which is what is inside the absolute value sign) is not. – Arthur Feb 10 '14 at 14:29
So I have to treat it like $||x²|-1|$ – depecheSoul Feb 10 '14 at 14:30

Factor $x^2 - 1$ and add $2$ on both sides to obtain

$|(x+1)(x-1)| \geq 2(x+1)$.

Now there are two cases. First, if $x \leq -1$, the RHS $2(x+1)$ is non-positive and hence the inequality is definitely fulfilled.

Second, let $x > -1$. But then, it follows that $|(x+1)(x-1)| = (x+1)|x-1|$, since $(x+1)$ is positive.

The inequality thus boils down to $(x+1)|x-1| \geq 2(x+1)$.

Eliminating the positive (by assumption of $x > -1$) factor $(x+1)$ then yields

$|x-1| \geq 2$.

This is either the case if $x-1 \geq 2$ ($\implies x \geq 3$), or $x-1 \leq -2$ ($\implies x\leq -1$). The second contradicts our assumption of $x > -1$, so $x\geq 3$ remains.

We conclude that the inequality is fulfilled if either (a) $x \leq -1$ or (b) $x \geq 3$.

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How did you get x<=-1 and x>-1. Did you use x²-1=0; x²=1; x=+-1; Thanks. But the you have +1 also – depecheSoul Feb 10 '14 at 15:03
Do you mean the two cases? This follows directly from the right hand side - it will be negative for $x < -1$ and positive for $x \geq -1$. Hence, for $x < -1$, the inequality is automatically fulfilled. – Martin Feb 10 '14 at 15:17

Another way this can be written:

One case is

$$(x^2 - 1) \ - \ 2 \ \ge \ 2x \ \ \Rightarrow \ \ x^2 \ - \ 2x \ - \ 3 \ \ge \ 0 \ \ \Rightarrow \ \ (x + 1) \ (x - 3 ) \ \ge \ 0$$

and the other is

$$(1 - x^2) \ - \ 2 \ \ge \ 2x \ \ \Rightarrow \ \ 0 \ \ge \ x^2 \ + \ 2x \ + \ 1 \ \ \Rightarrow \ \ 0 \ \ge \ (x + 1)^2 \ \ .$$

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$$|x^2 - 1| -2\ge 2x \iff |x^2 - 1| \ge 2x + 2 = 2(x + 1)$$ $$\iff |(x - 1)(x + 1)|\ge 2(x+1)$$

Now consider the cases $x^2\geq 1$ and $x^2 \lt 1$.

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does the case $x^2\ge1$ need to be treated as two sub-cases, $x\le1$ and $x\ge1$? – TooTone Feb 10 '14 at 14:44
For all (and only) those $x$ such that $-1\lt x \lt 1$, we have $x^2 \lt 1$ – amWhy Feb 10 '14 at 14:58