Verify and Improve: $|x+\frac{1}{x}|\geq 6$ Exercise:

Determine the intervals in which the following inequality is satisfied:
  $$|x+\frac{1}{x}| \geq 6$$


Attempt:
I apologize for not MathJax-ifying this; the following easy-to-read formatting is beyond my expertise.


Request:
Can I improve upon the solution to make it more elegant and concise?
 A: My answer tries to explain more precisely the symmetry involved.
Let's consider:
$$
f(x) \equiv x+\frac{1}{x}=-f(-x)
$$
Hence, the LHS of the inequality is an odd function of x. So, under an absolute value:
$$
|f(x)|=|f(-x)|\implies |f(x)|=|f(|x|)| \,\,\,\forall x
$$
Hence, you can replace $x\to|x|$ and solve accordingly as other user have suggested.
A: Possibly more concise approach (though yours is certainly sound):


*

*For $|x+\frac{1}{x}| \geq 6$, we need only consider $x>0$ by symmetry (exclude $x=0$ as you say)

*$x+\frac{1}{x}\ge6\implies x^2-6x+1\ge0\iff(x-3)^2\ge8\iff|x-3|\ge2\sqrt{2}$

*So, for $x>0$, our solution is $0<x\le3-2\sqrt2\cup x\ge3+2\sqrt2$

*For $x<0$, symmetry gives $2\sqrt2-3<x<0\cup x<-3-2\sqrt2$


*

*Alternatively, we can just replace every $x$ with $|x|$ to arrive directly at:



$$|x+\frac{1}{x}| \geq 6 \iff 0<|x|\le3-2\sqrt2\cup |x|\ge3+2\sqrt2$$
A: Square the inequality to get rid of the absolute value.
$$\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2\ge36.$$
s $x^2>0$, this can be rewritten
$$x^4-34x^2+1\ge0$$ and by completing the square,
$$(x^2-17)^2\ge17^2-1=2\cdot12^2.$$
Then
$$x^2-17\le-12\sqrt2\ \lor\ 12\sqrt2\le x^2-17,$$ 
$$x^2\le17-12\sqrt2\ \lor\ 12\sqrt2+17\le x^2$$ 
 and
$$|x|\le\sqrt{17-12\sqrt2}\ \lor\ \sqrt{12\sqrt2+17}\le|x|,$$
also written, after radical denesting
$$|x|\le3-2\sqrt2\ \lor\ 3+2\sqrt2\le|x|.$$
A: First of all you don't need to check for negative/positive $x$ values separately. Because if the inequality holds for some positive (or negative) $x$, then it holds for $-x$, too. So you just need to check it for one side.
Assuming $x>0$, we have
$$
\begin{align}
6&\leq x+\frac1x\\
0&\leq x^2-6x+1\\
0&\leq (x-3)^2-8\\
8&\leq (x-3)^2\\
2\sqrt{2}&\leq|x-3|\\
\end{align}
$$
which is equivalent to
$$2\sqrt{2}\leq x-3 \quad\mathrm{or}\quad -2\sqrt{2}\geq x-3\\$$
or
$$x\geq3+2\sqrt{2} \quad\mathrm{or}\quad 0<x\leq 3-2\sqrt{2}\\$$
Due to symmetry around zero (as I've mentioned in the beginning) the final answer is $$x\in(-\infty,-3-2\sqrt{2}]\cup[-3+2\sqrt{2},0)\cup(0,3-2\sqrt{2}]\cup[3+2\sqrt{2},\infty)$$
