How to solve $|1+1/x| > 2$ I am having a problem with this question:

Find all real $x$ that satisfies $|1 + 1/x| > 2$.

This is clearly not defined in $x = 0$.
By my logic, it should be solved with:

$1+1/x > 2$   or  $1+1/x < -2$

But the result I am getting from this is wrong. ($x<1$)
Correct result is 

$-1/3 < x < 1$

How to solve a problem like this? Why is this logic not working here:

$|x| > 2$
$x > 2$ or $x < -2$

 A: I suspect you failed "switch" signs for negative values.
Either $1 + \frac 1x < -2$ or $1 + \frac 1x > 2$.
If $1 +\frac 1x <-2$ then $\frac 1x < -3$ and $x$ is negative.
So $1 > -3x$ (inequality flipped because $x < 0$)
$-\frac 13 < x$ (ditto for $-3$).  And as $x$ is negative.
$-\frac 13 < x < 0$
OR
$1 +\frac 1x > 2$ so $\frac 1x > 1$ and $x$ is positive and $x < 1$.  So $0 < x < 1$.
So $-\frac 13 < x < 0$ or $0 < x  <1$ or $x \in (-\frac 13,0) \cup (0,1)$.
A: Taking from $$|1+\frac1x|>2\implies 1+\frac1x>2\ \text{or}\ 1+\frac1x<-2,$$  this gives us two cases:  
Case 1:  \begin{align} 1+\frac1x&>2\\ \frac1x-1&>0\\
\frac{1-x}x&>0\implies 0<x<1
\end{align}
Case 2:  \begin{align} 1+\frac1x&<-2\\ 3+\frac1x&<0\\
\frac{3x+1}{x}&<0\implies-\frac13<x<0
\end{align}
We can then take the union of the two sets, and the final answer is $\boxed{-\dfrac13<x<1, x\neq0}$.  Your mistake probably comes from solving the rational inequality incorrectly.

Edit:  here is how I would solve rational inequalities, in general.
Let $f(x)=\dfrac{3x+1}{x}$.  The zeroes of the numerator and denominator are $-\dfrac13$ and $0$.  Use these to form bounds of intervals.
\begin{array}{|r|c|c|c|}
\hline & 3x+1 & x & \text{sign of } f(x) \\
\hline
x<-\frac13 & - & - & +\\
-\frac13<x<0 & + & - & -\\
x>0 & + & + & + \\
\hline
\end{array}
Since we want $f(x)=\dfrac{3x+1}{x}<0$, it follows that the solution is $-\dfrac13<x<0$ from the table above.
A: If you square both sides, you obtain $$\left(1+1/x\right)^2>2^2,$$ or $$(1+1/x)^2-2^2>0,$$ which factors as $$(1+1/x-2)(1+1/x+2)>0,$$ or $$(1/x-1)(1/x+3)>0.$$ This implies $1/x<-3$ or $1/x>1.$ This implies $0>x>-1/3$ or $0<x<1.$ Hence the solution is the set of all $x\ne 0$ in $(-1/3,1).$
A: We have
$$1+\frac1x > 2 \iff \frac1x > 1 \iff 0 < x < 1$$
and similarly
$$1+\frac1x < -2 \iff \frac1x < -3 \iff -\frac13 < x < 0$$
Therefore we have
$$\left|1+\frac1x\right| > 2 \iff 1+\frac1x > 2 \text{ or } 1+\frac1x < -2  \iff -\frac13 < x < 0 \text{ or } 0 < x < 1$$
which is equivalent to $$x \in \left\langle -\frac13, 1\right\rangle\setminus \{0\}.$$
A: Let's follow your logic and see where it leads us. Start with the first case
$$1 + \frac{1}{x} > 2.$$
This means that $\frac{1}{x} > 2 - 1 = 1 \implies 0 < x < 1.$ Note that you got $x < 1,$ which is not true, because if $x < 0,$ we clearly have that $\frac{1}{x} < 0 < 1.$
On the other hand, your second case has
$$1 + \frac{1}{x} < -2.$$
In this case, $\frac{1}{x} < -2 - 1 = -3.$ Note that in this case, $x$ must be negative. Solving the inequality as such, we have that $0 > x > -\frac{1}{3}.$
We take the union of these two cases for the complete solution set: $-\frac{1}{3} < x < 1, x \ne 0.$
A: $x\not =0$;
1) $1+1/x >2$;
2) $1+1/x <-2$;
1) $1/x >1$;
Then $x >0$: 
$1>x$, i.e $0<x<1.$
2) $1/x < -3$;
Then $x<0$: 
$1>-3x$ ; $-1/3 <x$ ; i.e $-1/3<x<0$;
3) Altogether: $x \in (-1/3,0)\cup (0,1)$.
A: \begin{align}
|1+1/x|>2&\iff 1+1/x>2 \text{ or } (1+1/x)<-2\\
&\iff 0<x<1 \text{ or } -\frac13<x<0.
\end{align}
A: Alternatively, $x\ne 0$ and:
$$|x+1|>2|x| \iff (x+1)^2>(2x)^2 \iff (x-1)(3x+1)<0 \iff -1/3<x<1$$
Hence:
$$x\in (-1/3,0)\cup (0,1)$$
A: If you divide it by .2 or .5 or basically values < 1 you get higher values:
1 + (1 / .2) = 6
1 + (1 / .5) = 3 

I'm not a mathematician so this may be missing the point of your equation, but this does make it true I believe since the order of operation is division before addition
