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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?

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    Well, not quite. Nothing can be both $\geq 2$ and $\leq -2$. Instead, *one* of those must be true. – 2017-01-12
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    @ozk How can -2 be greater than 2? – 2017-01-12
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    Imagine that one says my weight is greater than 60 and lower than 50! – 2017-01-12
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    [Duplicate(ish)?](http://math.stackexchange.com/questions/2082597/verify-and-improve-x-frac1x-geq-6) – 2017-01-12
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    By any chance, are you using Courant's book? I just solved this same exercise from it a couple days ago. – 2017-01-12

4 Answers 4

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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 $x\ne0$

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    but our inequality $$\left |x+\frac{1}{x}\right|\geq 2$$ makes in the case $x=0$ no sence, i meant the inequalties are equivalent!!!! – 2017-01-12
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    As it makes no sense it is important to acknowledge. – 2017-01-12
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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.

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Let $x>0$. Use AM-GM $$x+\frac1x\ge2\sqrt{x\cdot\frac1x}=2$$ Let $x<0$. Then $y=-x>0$ $$|x+\frac1x|=-x-\frac1x=y+\frac1y\ge2 \Rightarrow -x-\frac1x\ge2\Rightarrow x+\frac1x\le-2$$

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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?