Is there a lower bound for the following rational expression, in terms of $b^2$? In the following, $b$ and $r$ are both positive integers.
Is there a lower bound for the following rational expression, in terms of $b^2$?
$$L = \frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + 1)}$$
What I do know is that (a crude upper bound is) $L < b^2$.  My question:  Is it possible to get a lower bound for $L$ in terms of $b^2$ alone?
Edit:  Here is an attempt to derive a (crude) lower bound for $L$, using no more than $r \geq 1$ and the AM-GM Inequality.
Since $r \geq 1$, the numerator of $L$ is
$$b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right) \geq 4b^4 + 4b^2 = 4{b^2}(b^2 + 1).$$
Now, for the denominator of $L$, we use the AM-GM Inequality to get
$$\left(2^{r+1} - 1\right)(2^r b^2 + 1) \leq \frac{\left(2^{r+1} + 2^r b^2\right)^2}{4}.$$
Simplifying, we obtain
$$\left(2^{r+1} - 1\right)(2^r b^2 + 1) \leq \left(2^r + 2^{r-1} b^2\right)^2 < 2^{2r}(b^2 + 1)^2.$$
Consequently, I finally have:
$$L > \frac{2^{2-2r}\cdot{b^2}}{b^2 + 1}$$
which is still expressed in terms of the extra variable $r$.
I was wondering if anybody out there has any bright ideas on how to eliminate $r$ from a (hopefully) nontrivial lower bound for $L$, in terms of $b^2$.
Thanks!
 A: $$\frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + 1)}=\frac{2^{r+1}(2^r-1)b^4 + (3\cdot 2^r - 2)b^2}{\left(2^{r+1} - 1\right)2^r(b^2 + 2^{-r})}=\frac{b^4 + \frac{3\cdot 2^r - 2}{2^{r+1}(2^r-1)}b^2}{\left(1 +\frac{2^r}{2^r-1}\right)(\frac{1}{2}b^2 + 2^{-r-1})}\geq b^2\frac{b^2+3\cdot 2^{-r-1}- \frac{1}{2^{r}(2^r-1)}}{b^2+2^{-r}}\geq b^2\frac{b^2+2^{-r}+2^{-r-1}-2^{-2r+1}}{b^2+2^{-r}}\geq b^2$$
A: Taking off from Matt's answer, I get:
$$\frac{1}{2}\left(1 + \frac{2^r}{2^r - 1}\right)(b^2 + 2^{-r}).$$
We need an upper bound for $\frac{1}{2}\left(1 + \frac{2^r}{2^r - 1}\right)$.
We have the upper bound:
$$\frac{1}{2}\left(1 + \frac{2^r}{2^r - 1}\right) = \frac{1}{2}\left(2 + \frac{1}{2^r - 1}\right) \leq \frac{3}{2}$$
since $r$ is a positive integer (and hence we can take $r \geq 1$).
Consequently, we now have the (adjusted) lower bound:
$$\frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + 1)}=\frac{2^{r+1}(2^r-1)b^4 + (3\cdot 2^r - 2)b^2}{\left(2^{r+1} - 1\right)2^r(b^2 + 2^{-r})}=\frac{b^4 + \frac{3\cdot 2^r - 2}{2^{r+1}(2^r-1)}b^2}{\left(1 +\frac{2^r}{2^r-1}\right)(\frac{1}{2}b^2 + 2^{-r-1})}\geq \left(\frac{2}{3}{b^2}\right)\frac{b^2+3\cdot 2^{-r-1}- \frac{1}{2^{r}(2^r-1)}}{b^2+2^{-r}}\geq \left(\frac{2}{3}{b^2}\right)\frac{b^2+2^{-r}+2^{-r-1}-2^{-2r+1}}{b^2+2^{-r}}\geq \frac{2b^2}{3}.$$
I hope what I have written is correct!  =)
