inequality
How come since $e>1 \implies e^{-1/2} < 1^{-1/2}$. I know that one reverses the inequality signs when we take reciprocal of both sides or multiplies by a negative number. I have never seen inequality sign inverted because we raise both sides to the negative power.
What would in this case then, given:
$a > -\ln (2e^{-1/2}-1)$
and I wanted to take exponent and as a power have each side of the inequality:
$e^a \ \ ? \ \ e^{\ln(2e^{-1/2}-1)^{-1}}$
Dividing by a negative number is the same as dividing by a positive number and then multiplying by $-1$. Dividing an inequality by a positive number retains the same inequality. But, multiplying by $-1$ is the same as switching the signs of the numbers on both sides of the inequality, which reverses the inequality: $$ \tag{1} a\lt b\quad\iff -a\gt -b. $$ You should be able to convince yourself why the above is true by looking at the number line and considering the various cases involved.
Seeing why (1) is true is not too hard.
Here is the hand waving approach I suggested above:
Consider, for example, in (1), the case when $a$ is negative and $b$ is positive. We have $a<b$. Then $-a$ is positive and $-b$ is negative. Thus, we have $-b<-a$.
As another case, suppose $a$ and $b$ are both negative with $a<b$. Switching the signs here makes the resulting numbers both positive with $-a>-b$ (you can see this by drawing the points on the number line and noting that with the given conditions, $b$ is closer to the origin than $a$):
).
The other cases can be handled similarly.
But, perhaps a bit of rigor is needed here.
Recall that
$$a<b\quad\text{ if and only if }\quad b-a\quad \text{ is positive.}$$
Now, $b-a$ is positive if and only if $(-a)-(-b) =-a+b=b-a$ is positive. So
$a<b$ if and only if $-a> -b$.
It's not entirely true that when you take the reciprocals of both sides, you change "$\lt$" to "$\gt$" and vice-versa. If $a$ and $b$ are both positive or if they are both negative, and $a<b$, then $1/a>1/b$. But if one is positive and the other negative, and $a<b$ (which means $a$ must be the one that's negative) then $1/a<1/b$; the direction doesn't get reversed.
Generally if $a<b$ and $g$ is a strictly decreasing function, then $g(a)>g(b)$. That's actually the definition of the concept of "strictly decreasing function". So the fact that when you multiply by a negative number, you invert the inequality relation, is the same as saying that multiplication by a negative number is a strictly decreasing function. For example, if $g(x) = -5x$ for all values of $x$, then $g$ is a strictly decreasing function. If $g(x)=1/x$, then the restriction of $g$ to the positive numbers is a strictly decreasing function, and the restriction of $g$ to the negative numbers is a strictly decreasing function, but $g$, over its whole domain, is not a strictly decreasing function.
$\arccos$ is a strictly decreasing function: as a number increases from $-1$ to $1$, its arccosine decreases, i.e. if $a<b$, then $\arccos a>\arccos b$.
One thing that will tell you that a function is strictly decreasing is that it's derivative is everywhere negative and its domain has no gaps. The reciprocal function has an everywhere negative derivative, but its domain has a gap at $0$.
Best Answer
Taking the reciprocal of each side (which is the same thing as raising to the negative first power) only flips the inequality if $a\times b$ is positive. That is, it only holds if $a,b$ are of the same sign.
Consider $a>b$.
If $ab>0$, then dividing both sides above by $ab$ gets $\frac1b>\frac1a$ as per the rule. The inequality appears to have flipped, $a^{-1} < b^{-1}$.
However, if $ab<0$, then dividing both sides above by $ab$ gets $\frac1b<\frac1a$, or in other phrasing, $a^{-1} > b^{-1}$.