Csdrhrt

Suppose we want to solve the following inequality

$ frac{1}{x} + 1 >0 $

So we proceed as follows

$ frac{1}{x} > -1 $

$ x < -1 $

Thus $ x in (-infty, -1)$

But the solution to the inequality is

$ Re - [-1, 0) $

The problem is that $$1/x >−1$$ does not imply $$ x<-1$$

For example for $x=5$ we have $1/5 >-1$ but $5>-1$

One has to be careful with inequalities involving variables.

To avoid case analysis write
$$
frac{1}{x}+1>0ifffrac{1+x}{x}>0ifffrac{x(1+x)}{x^2}>0.
$$
From here we see that $x(1+x)>0$ i.e. $x>0$ or $x<-1$.

The problem is the sign of $x$. When $x<0$ you must flip the direction of the inequality when multiplying through.

There are two cases. Firstly for $x>0$;
begin{align*}
&
frac{1}{x}>-1\
Rightarrow & 1>-x\
Rightarrow & -1<x\
Rightarrow & x>-1\
end{align*}

So $x>0$ and $x>-1$, together these conditions imply that $x>0$.

Then for $x<0$;

begin{align*}
&
frac{1}{x}>-1\
Rightarrow & 1<-x\
Rightarrow & -1>x\
Rightarrow & x<-1\
end{align*}

Putting the two cases together we have;

$$xin(-infty,-1)cup(0,infty)=mathbb{R}setminus[-1,0]$$

EDIT: Note that your given solution includes $x=0$, but mine doesn't. As $xrightarrow 0$, the inequality holds, but it is undefined at zero itself which is why I have exluded this point.

That's a tricky one.

Here is a possible solution. Clearly $xne -1$ follows from first inequality. Now from 2nd to 3rd inequality note that $frac{1}{x}>-1implies -x<1$ when $x$ is nonnegative i.e. $x$ can be zero or any positive number i.e. $xin [0,infty)$, again $frac{1}{x}>-1implies 1<-x$ when $x$ is any negative number such that $xne 0$ and $xnotin [-1,0]$ if so then it will contradict the last inequality. Hence, $xin (-infty,-1)$ . Thus in general we have: $xinmathbb{R}-[-1,0)$.

Hope that works.