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?
 A: Note that $x$ cannot be negative as $|x+1|$ is always nonnegative.
With that mind, observe that $|x+1| = x+1, \forall x > 0$. Therefore
$$ \frac{1}{x+1} < \frac{1}{2x} $$
which results in $x < 1$ as you have done.
So the final result is $x \in (0,1)$
A: The left hand side of $$\frac{1}{|x+1|} < \frac{1}{2x}$$ is positive so the right hand side must be positive. Therefore $x>0$ which implies $x+1>0$ and $|x+1|=x+1.$
Upon substitution, the inequality becomes  $$\frac{1}{x+1} < \frac{1}{2x}$$ with the condition $x>0.$ The solution is $0<x<1$. 
A: First observe this inequation is defined for $x\ne 0,-1$ and that it implies $x>0$. Now you can square both sides to remoave the absolute value:
$$\frac1{|x+1|}<\frac1{2x}\iff|x+1|>2x\iff(x+1)^2>4x^2\iff 0>3x^2-2x-1.$$
One obvious root of the quadratic polynomial is $x=1$, and  the other root is negative since their product is $-\dfrac13$.
So $3x^2-2x-1<0\iff -\dfrac13<x<1$. Taking into account the condition on$x$, we obtain
$$0<x<1.$$
A: 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 $$(\infty,-1)\cap\left(-\infty,-\frac13\right)=(-\infty,-1)$$
