Prove that $\frac{d}{2^m + 1} + \frac{m}{2}\frac{2^m}{2^m+1} \geq \frac{\log_2 d}{4}$ I am reading Tau and Vu's book Additive Combinatorics, and I came across a step in a proof that I am not able to verify.
On page 252, in the last line of the proof of Theorem 6.4, it is stated that the following inequality holds for all integers $m \geq 1$ and $d \geq 16$:
$\frac{d}{2^m + 1} + \frac{m}{2}\frac{2^m}{2^m+1} \geq \frac{\log_2 d}{4}$
I would appreciate a proof of this fact. I have tried various ways of rearranging the inequality, and I have tried fixing one variable and then taking the derivative with respect to the other, but I'm not sure if this will work. Any help would be appreciated.
Note: As I state in the comments below, we may assume $m \leq d$. I'm not sure if this is necessary though.
The result in the book is cited to be in a paper by Shearer, which I've found here:
https://www.sciencedirect.com/science/article/pii/0012365X8390273X, But I can't find this result in this paper. Am I just missing it?
 A: There are a number of routes to establish this inequality, in part because it's true with plenty of room to spare (one can get a feel for this by thinking about the cases when $d$ is massive and $m$ fixed, or vice versa, or $d=m$). Below is one possible explicit demonstration. 
Let $k\geq 4$ be such that $2^k \leq d< 2^{k+1}$. It is enough to show that for all $m\geq 1$ and $k\geq 4$
$$ \frac{2^k}{2^m+1}+\frac{m}{2}\frac{2^m}{2^m+1} \geq \frac{k+1}{4},$$
or, after rearranging,
$$ 2^{k+2}+ m2^{m+1} \geq (k+1)(2^m+1).$$
If I rewrite this as
$$2^{k+2} \geq (k+1-2m)2^m + k+1$$
then it is clear that it holds if $2m\geq k +1$, for then it follows from $k+1\leq 2^{k+2}$. Otherwise, $m\leq k/2$, and so 
$$ (k+1)(2^m+1) \leq (k+1)(2^{k/2}+1)\leq (2^{k/2}+1)^2\leq 2^{k+2} $$
where I have used $k\leq 2^{k/2}$ (valid for $k\geq 4$) and $2^{k/2}+1\leq 2^{k/2+1}$.
A: It is easy to check that for each real $\alpha$ we have $2^{\alpha+1}\ge\alpha$. Now put $d=2^{m+\alpha}$. Then 
$$\frac{d}{2^m + 1} + \frac{m}{2}\frac{2^m}{2^m+1}\ge \frac{2^{m+\alpha}+\frac{m}{2}2^{m}}{2^{m+1}}=2^{\alpha-1}+\frac{m}4\ge 
\frac{m+\alpha}{4}=\frac{\log_2 d}{4}.$$
