<s>[INST] 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!
 [/INST] Shortly $${ \frac { 1 }{ 1+x }  }>-1\\ \frac { 1 }{ 1+x } +1>0\\ \frac { x+2 }{ x+1 } >0\\ \frac { \left( x+1 \right) \left( x+2 \right)  }{ { \left( x+1 \right)  }^{ 2 } } >0\\ \left( x+1 \right) \left( x+2 \right) >0\\ x\in \left( -\infty ;-2 \right) \cup \left( -1;+\infty  \right) \\ $$
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