Inequality solution with absolute value I have the following inequality:
$$ \left\rvert x + \frac{1}{x} \right\lvert \ge 2$$
can I say that the distance of the expression inside the absolute value from $0$ is greater or equal to 2 (from the positive side of the real axis) and smaller or equal to $-2$ (from the negative side) so:
$$ \left\rvert x + \frac{1}{x} \right\lvert \ge 2 \implies -2\ge x + \frac{1}{x} \ge 2$$
can I use it like that to solve it? 
 A: This line is incorrect:
$$\rvert x + \frac{1}{x} \lvert \ge 2 \implies -2\ge x + \frac{1}{x} \ge 2$$
The absolute value means the value inside could have been positive or negative so it actually becomes:
$$\pm\left(x+\frac{1}{x}\right)\ge2$$
When you go to move the negative sign to the other side you must split this into two separate equations:
$$x+\frac{1}{x}\ge2\text{ and }x+\frac{1}{x}\le-2$$
You can then use these two inequalities to finish solving the actually problem (however there are other easier ways to solve it).
First inequality:
$$x+\frac{1}{x}-2\ge0$$
$$\frac{x^2-2x+1}{x}\ge0$$
$$\frac{(x-1)^2}{x}\ge0$$
The numerator is always positive so we require $x>0$. Note we can not have $x=0$ as we can not divide by 0.
Second inequality:
$$x+\frac{1}{x}+2\le0$$
$$\frac{x^2+2x+1}{x}\le0$$
$$\frac{(x+1)^2}{x}\le0$$
The numerator is always positive so we require $x<0$. Note we can not have $x=0$ as we can not divide by 0.
Combined solution
We can have either $x>0$ or $x<0$ so the solution is $x\ne0$
A: You could also approach this from a slightly different angle: using the fact that $|x|^2 = x^2$, square both sides of the original inequality, and work with that.
A: Let $x>0$. Use AM-GM
$$x+\frac1x\ge2\sqrt{x\cdot\frac1x}=2$$
Let $x<0$. Then $y=-x>0$
$$|x+\frac1x|=-x-\frac1x=y+\frac1y\ge2 \Rightarrow -x-\frac1x\ge2\Rightarrow x+\frac1x\le-2$$
A: A geometric approach: Consider a square with dimensions $x+1/x$ and calculate the area of the four regions of this square. These regions are $x^2$,$1/x^2$ and $2$. Conclusion?
