Inversion of the inequality sign when raising to a negative power 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:
$e^a \ \ ? \ \ e^{\ln(2e^{-1/2}-1)^{-1}}$

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, $a^{-1} > b^{-1}$.
