Solving a rational inequality with an unknown using cases So the question is,

For what values of x is $\dfrac{1}{1 + x}> -1 $ ?

Now, one way to do it is write the inequality relative to zero, and then find the behavior of the graph relative to the zeroes of the numerator and denominator:
$\dfrac{1}{1 + x}> -1 \implies \dfrac{1}{1+x} + 1> 0 \implies \dfrac{1}{1+x} + \dfrac{1+x}{1+x} > 0 $
$\implies \dfrac{2+x}{1+x} > 0$, and we simply find what happens when $x < -2, -2 < x < -1$, and $x > -1$.
What I don't understand is why I can't answer this by multiplying through by $1 +x$ if I consider both a) $1 +x >0$ and b) $1+x < 0$.
Here's the attempt: 
Case a): 
$1 + x > 0 \implies x > -1$, so 
$\dfrac{1}{1+x} > -1$
$\implies (1+x) \dfrac{1}{1+x} > -1 (1+x) $ (we can multiply both sides by 1+x because we assume 1+x is positive)
$\implies x > -2 $.
Hence, the inequality is true when $x > -2$ and $x > -1$--so $x > -1$.
Case b): $1+x < 0 \implies x < -1$, so 
$\dfrac{1}{1+x} > -1$
$\implies (1+x) \dfrac{1}{1+x} < -1 (1+x) $ (we assume $1+x$ is negative, so we reverse the sign)
$\implies 1 < 1 + x$
$\implies x > 0$.
Hence, the inequality is true when $x < -1$ and $x > 0$--this is never true.
So the real answer is $x > -1$ or $x <-2$, and I think the reason my attempt at multiplying through by an unknown doesn't work is that, in my assumption $x\in (-1, \infty)$ clearly includes both negative and positive values, and that leads the answer to be wrong.
But I'm not 100 percent certain it's impossible, and would like some help!
 A: Shortly $${ \frac { 1 }{ 1+x }  }>-1\\ \frac { 1 }{ 1+x } +1>0\\ \frac { x+2 }{ x+1 } >0\\ \frac { \left( x+1 \right) \left( x+2 \right)  }{ { \left( x+1 \right)  }^{ 2 } } >0\\ \left( x+1 \right) \left( x+2 \right) >0\\ x\in \left( -\infty ;-2 \right) \cup \left( -1;+\infty  \right) \\ $$
A: I think the question has been well-answered, but in case you were looking for a way to multiply through and not have to worry about these cases, try this:
$$\frac{1}{1+x} > -1$$ Multiplying through by the positive quantity $(1+x)^2$ gives, $$ (1+x)^2 \frac{1}{1+x} > -(1+x)^2$$ $$ 1+ x > -(1+x)^2$$ $$(1+x) + (1+x)^2 > 0 $$$$ (1+x) [1+ (1+x)]>0$$ $$(1+x) (2+x)>0$$ which has solutions of $x<-2$ or $x>-1$. 
A: The following specifically answers this part of OP's question:

What I don't understand is why I can't answer this by multiplying through by $1 +x$ if I consider both a) $1 +x >0$ and b) $1+x < 0$.

This does in fact work, and the posted attempt failed only because of a simple mistake in case b).

Case b): $1+x < 0 \implies x < -1$, so 
  $\dfrac{1}{1+x} > -1$
  $\implies (1+x) \dfrac{1}{1+x} < -1 (1+x) $ (we assume $1+x$ is negative, so we reverse the sign)
  $\color{red}{\implies 1 < 1 + x}\;\;\cdots$

The part in red is wrong. After multiplication, the inequality is $\,1 \lt \color{red}{-1 -x} \iff x \lt -2\,$, instead. Once this is corrected, the rest of the proof works, and gives the correct end result.
The problem appears to be with this comment "we assume $1+x$ is negative, so we reverse the sign" which I am not sure what means. When multiplying $-1 \cdot (1+x)$ the result is always $-1 \cdot (1+x)$ regardless of the sign of $1+x\,$. What does change when $1+x$ is negative is that the direction of the inequality gets reversed from $\,\gt\,$ to $\,\lt\,$ (which part was done correctly in the post).
