TITLE: Inequality; when we switch signs? From basic to complicated
QUESTION [10 upvotes]: Let's say I have
$$a > b$$
Now we know that the inequality sign will switch if:

We take the reciprocal of both sides
Multiply $-1$ to both sides.

But, are there more? What if we do more "complicated" things to it? 
Will it change if I apply inverse trig on it? For instance is it still true then if 
$$\arccos(a) > \arccos(b)$$
Note that $a,b \in\mathbb R$.
So what I really want to ask is, are there any other operations that would make you change the inequality signs?

REPLY [4 votes]: Multiplication by negative numbers flips inequalities. Multiplication by positive numbers respects inequalities.
If $a\lt b$ and $c\lt 0$, then $ac\gt bc$; if $d\gt 0$, then $ad\lt bd$. 
You can deduce the case of reciprocals (which is more complicated than what you wrote) from this. 
If $0\lt a\lt b$, then multiplying by $\frac{1}{ab}$ (which is positive) gives $\frac{a}{ab} \lt \frac{b}{ab}$, which yields $\frac{1}{b}\lt \frac{1}{a}$; i.e., a "flipped inequality".
If $a\lt b \lt 0$, then multiplying by $\frac{1}{ab}$ (which is also positive, since $a$ and $b$ are both negative) again yields $\frac{1}{b}\lt \frac{1}{a}$.
If $a\lt 0 \lt b$, then multiplying by $\frac{1}{ab}$ (which is negative) yields
$$\frac{a}{ab}\gt \frac{b}{ab}$$
or $\frac{1}{b}\gt \frac{1}{a}$; that is, the inequality is not "flipped."
A function $f$ is strictly increasing if $a\lt b$, both in the domain of $f$, implies $f(a)\lt f(b)$. Examples of increasing functions are $f(x)=x^3$, $f(x)=\arctan(x)$, $f(x)=e^x$, $f(x)=\log(x)$. A function $f$ is increasing if $a\lt b$, both in the domain of $f$, implies $f(a)\leq f(b)$.
Symmetrically, a function $f$ is strictly decreasing if $a\lt b$ and both in the domain imply $f(a)\gt f(b)$; and decreasing if $a\lt b$ and both in the domain implies $f(a)\geq f(b)$.
You can also deduce most of the case of reciprocals from this: $f(x)=\frac{1}{x}$ is strictly decreasing on the positive reals, so if $0\lt a \lt b$, then $f(a)\gt f(b)$; $g(x)=\frac{1}{x}$ is strictly decreasing on the negative numbers, so if $a\lt b\lt 0$, then $g(a)\gt g(b)$. But the case $a\lt 0\lt b$ has to be dealt with separately anyway.