solving $1+\frac{1}{x} \gt 0$ In solving a larger problem, I ran into the following inequality which I must solve:
$$ 1+\frac{1}{x} \gt 0.$$ 
Looking at it for a while, I found that $x\gt 0$ and $x\lt -1$ are solutions. Please how do I formally show that these are indeed the solutions. 
 A: After reading your comment here is how to go about it when you multiply by $x^2$. So we need to solve 
$$x^2 + x > 0.$$
That can be written as
$$x(x + 1) > 0.$$
Now the product of two numbers is positive when both numbers are positive or when both are negative. Therefore we have two cases to solve:
Case 1:
$x >0$ and $x + 1> 0$ 
Both of these inequalities are satisfied when $x > 0$.
Case 2:
$x < 0$ and $x + 1 < 0$
Both of these inequalities are satisfied when $x < -1$.
Hope that helps!
A: note that $\frac{x+1}{x}>0$ and $(x+1)x>0$ have the same solution(or, as you mentioned, multiply them both with $x^2$)
generally speaking, bringing the denominator over the bar doesn't change the answer, but be careful that the denominator can't be zero.
A: There are two cases.  If $x>0$ we may multiply both sides by $x$ and the inequality is unchanged, i.e. $x+(1/x)x>0(x)$, which simplifies to $x+1>0$ or $x>-1$.  This case is the combination of $x>0$ and $x>-1$, which is $x>0$.
If $x<0$ we multiply both sides by $x$ and the inequality reverses, i.e. $x+(1/x)x<0(x)$, which simplifies to $x+1<0$ or $x<-1$.  This case is the combination of $x<0$ and $x<-1$, which is $x<-1$.
A: You asked how to show this formally, so a picture is possibly not very useful, but I include one because I need practice with tikz. Vadim's answer above is excellent.

