Is $\frac{(c-a)}{(c-b)}>\frac{\exp(-at)-\exp(-ct)}{\exp(-bt)-\exp(-ct)}$?

Question

Suppose, $$a<b<c$$ then is $$\frac{(c-a)}{(c-b)}>\frac{\exp(-at)-\exp(-ct)}{\exp(-bt)-\exp(-ct)}$$ for all $t>0$?

Answer 1

It is exactly the other way round which can be seen using the generalized mean value theorem (GMVT):

$$\frac{\exp(-at)-\exp(-ct)}{\exp(-bt)-\exp(-ct)}= \underbrace{\frac{\exp(-ct)}{\exp(-ct)}}_{=1}\cdot \frac{\exp((c-a)t)-1}{\exp((c-b)t)-1}$$ $$ \stackrel{GMVT: 0<\tau <t}{=}\frac{(c-a)\exp((c-a)\tau)}{(c-b)\exp((c-b)\tau)} = \frac{(c-a)}{(c-b)}\cdot\underbrace{\exp((b-a)\tau)}_{>1} \color{blue}{>} \frac{(c-a)}{(c-b)}$$