Solving $\frac{1}{|x+1|} < \frac{1}{2x}$ Solving $\frac{1}{|x+1|} < \frac{1}{2x}$
I'm having trouble with this inequality. If it was $\frac{1}{|x+1|} < \frac{1}{2}$, then:
If $x+1>0, x\neq0$, then 
$\frac{1}{(x+1)} < \frac{1}{2} \Rightarrow x+1 > 2 \Rightarrow x>1$
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2} \Rightarrow -(x+1) > 2 \Rightarrow x+1<-2 \Rightarrow x<-3$
So the solution is $ x \in (-\infty,-3) \cup (1,\infty)$
But when solving $\frac{1}{|x+1|} < \frac{1}{2x}$,
If $x+1>0, x\neq0$, then
$\frac{1}{x+1} < \frac{1}{2x} \Rightarrow x+1 > 2x \Rightarrow x<1 $
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2x} \Rightarrow -(x+1) > 2x \Rightarrow x+1 < -2x \Rightarrow x<-\frac{1}{3}$
But the solution should be $x \in (0,1)$.
I can see that there can't be negative values of $x$ in the inequality, because the left side would be positive and it can't be less than a negative number. But shouldn't this appear on my calculations?

So from first case where $x\in(-1,\infty)$, you get the permissible via of $x$ to be $$(-1,\infty)\cap(-\infty,1)=(-1,1)$$
In the second case where $x\in(-\infty,-1)$ you get the permissible via of $x$ to be $$(\infty,-1)\cap\left(-\infty,-\frac13\right)=(-\infty,-1)$$
