Thinking of a lower bound for a function

[TaPaKaH]

A function $f:\mathbb{R}^3_+\to[0,1]$ defined as $f(\lambda,\beta,x)=1-e^{-\frac{\lambda}{\beta}\left(1-e^{-\beta x}\right)}$ serves a lot of pain under integration.
As this function is used to describe a lower bound, could anyone suggest another non-zero function that would be smaller than $f$?

 

[mfb]

Making f smaller <-> making the outer exponential larger <-> making the argument larger <-> making the negative argument smaller <-> making the inner exponential larger
Hmm, bad direction.

You can use approximations like $e^{-x}<max\left(\frac{1}{e},1-\frac{x}{2}\right)$ for example (possible for both exponentials), but I'm not sure if that makes calculations easier.

 

[TaPaKaH]

Having one exponential is perfectly acceptable, it's the $\exp(...\exp(...))$ bit that's causing the integration problems.
It seems that $g(x)=(1-e^{-\frac{\lambda}{\beta}})\cdot\min(\frac{1}{e},1-e^{-\beta x})$ is a good enough lower bound. Thank you.

EDIT: Just found an even better bound: $f(\lambda,\beta,x)>(1-e^{-\frac{\lambda}{\beta}})(1-e^{-\beta x})$.

 

[mfb]

This is equivalent to $$1-q^r > (1-q)r$$ for 0<r,q<1. I don't find a counterexample, but I wonder why this is true (and how to show it).

($q=\exp(-\frac \lambda \beta)$ and $r=1-\exp(-\beta x)$)

 

[micromass]

This is equivalent to $$1-q^r > (1-q)r$$ for 0<r,q<1. I don't find a counterexample, but I wonder why this is true (and how to show it).

($q=\exp(-\frac \lambda \beta)$ and $r=1-\exp(-\beta x)$)

http://en.wikipedia.org/wiki/Bernouilli_inequality#Generalization

 

[mfb]

Ah nice. It needs some re-writing, but then it is really just Bernoulli's inequality. Thanks.