Solving $\frac{1}{|x+1|} < \frac{1}{2x}$ Solving $\frac{1}{|x+1|} &lt; \frac{1}{2x}$

I'm having trouble with this inequality. If it was $\frac{1}{|x+1|} &lt; \frac{1}{2}$, then:

If $x+1&gt;0, x\neq0$, then 

$\frac{1}{(x+1)} &lt; \frac{1}{2} \Rightarrow x+1 &gt; 2 \Rightarrow x&gt;1$

If $x+1&lt;0$, then

$\frac{1}{-(x+1)} &lt; \frac{1}{2} \Rightarrow -(x+1) &gt; 2 \Rightarrow x+1&lt;-2 \Rightarrow x&lt;-3$

So the solution is $ x \in (-\infty,-3) \cup (1,\infty)$

But when solving $\frac{1}{|x+1|} &lt; \frac{1}{2x}$,

If $x+1&gt;0, x\neq0$, then

$\frac{1}{x+1} &lt; \frac{1}{2x} \Rightarrow x+1 &gt; 2x \Rightarrow x&lt;1 $

If $x+1&lt;0$, then

$\frac{1}{-(x+1)} &lt; \frac{1}{2x} \Rightarrow -(x+1) &gt; 2x \Rightarrow x+1 &lt; -2x \Rightarrow x&lt;-\frac{1}{3}$

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?
Note that $x$ cannot be negative as $|x+1|$ is always nonnegative.

With that mind, observe that $|x+1| = x+1, \forall x &gt; 0$. Therefore

$$ \frac{1}{x+1} &lt; \frac{1}{2x} $$

which results in $x &lt; 1$ as you have done.

So the final result is $x \in (0,1)$
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 $$(\infty,-1)\cap\left(-\infty,-\frac13\right)=(-\infty,-1)$$