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?

• What are $m,d$? – Jair Taylor Jun 10 at 22:57
• Hi sorry - they are integers. – Tanny Libman Jun 10 at 23:27
• They have no graph-theoretic meaning? – Jair Taylor Jun 10 at 23:33
• $d$ is the max degree in a graph and $m$ is a subset of the set of neighbors of some vertex, so indeed $m \leq d$ and this may be helpful! – Tanny Libman Jun 11 at 18:52