TITLE: solving $1+\frac{1}{x} \gt 0$
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: 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!

REPLY [3 votes]: 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.