If you take the reciprocal in an inequality, would it change the $>/< $ signs? Example:$$-16<\frac{1}{x}-\frac{1}{4}<16$$ 
In the example above, if you take the reciprocal of $$\frac{1}{x}-\frac{1}{4} = \frac{x}{1}-\frac{4}{1}$$
would that flip the $<$  to $>$ or not?
In another words, if you take the reciprocal of $$-16<\frac{1}{x}-\frac{1}{4}<16$$   would it be like this: $$\frac{1}{-16}>\frac{x}{1}-\frac{4}{1}>\frac{1}{16}$$

If $a$ and $b$ have the same nonzero sign, then
$$a<b \iff \frac1a > \frac 1b$$
(i.e., taking reciprocals reverses the inequality).
If $a$ and $b$ have opposite (nonzero) signs, then
$$a<b\iff \frac1a <\frac1b$$ (i.e., taking reciprocals preserves the inequality).
These follow from the fact that the function $f(x)=1/x$ defined on the nonzero reals is strictly decreasing and sign-preserving on each component $(-\infty,0)$ and $(0,\infty)$ of its domain.
If either of $a$ or $b$ is zero, then you can't take reciprocals.
Finally, compound inequalities like $a<b<c$ should be separated into "$a<b$ and $b<c$" and each component considered separately.
I also remark that inverting a sum is not the same as inverting the addends separately.
