Suppose we have random variables $X_1$ and $X_2$ that are independently and identically distributed.
Suppose I am interested in the inequality $\frac{5}{X_1-X_2} > 10$.
How can I multiply both sides of this inequality by $X_1-X_2$
Specifically, since $X_1$ and $X_2$ are random variables, I do not know if $X_1 -X_2$ is positive or negative. Therefore, I don't know if I need to reverse the sign of the inequality or not.
An example of where I get confused. Consider the statement $$\frac{5}{X_1-X_2} > 10$$ IF AND ONLY IF $$.5 > X_1-X_2$$. This would not be a correct statement would it? In particular, the "IF" direction would fail, i think.
My logic is as follows:
First consider the "ONLY IF" direction. This direction holds because
if $\frac{5}{X_1-X_2} > 10$ then $X_1-X_2$ must be positive, because a negative number cannot be greater than $10$, and we get the result by rearranging.
Now for the "IF" direction, which does not hold.
Suppose $.5>X_1-X_2$. If $X_1>X_2$ then we can rearrange to get $\frac{5}{X_1-X_2} > 10$ by $X_1-X_2$ . However, we cannot be sure that $X_1>X_2$ and if $X_1<X_2$ then we have that $\frac{.5}{X_1-X_2}<10$ when we rearrange so the "IF" direction does not hold
I think $0.5>X_1−X_2$ will fail, like you say. Yes, it must be positive. If you consider $d=X_1-X_2$, then you can graph $\frac5d$ and $10$. The inequality will only be true when $0<d<0.5$ (which incorporates your "must be positive"), so you were almost right:
$$0<X_1-X_2<0.5$$