When I learn the book "Classical Fourier Analysis 3rd"(GTM249) by Loukas Grafakos, I encounter a difficulty. On page 90 line 3, the author write: $$\frac{6^{n}R^{n}}{(|x|+R)^{n}}+\sum_{k=0}^{\infty}\frac{1}{2^{nk}}\frac{6^{n}(2^{k+1}R)^{n}}{(|x|+2^{k+1}R)^{n}}\leq \frac{C_{n}\ \log(e+|x|/R)}{(1+|x|/R)^{n}} \,.$$ Then the author says:"where the last estimate follows by summing separately over k satisfying $2^{k+1}\leq |x|/R$ and $2^{k+1}\geq |x|/R$"
Frankly speaking, I really do not know how to prove the above inequality although the author write the hint. Thanks!
It clearly suffices to prove that $$\ \sum_{k=0}^{\infty}\frac{1}{2^{nk}}\frac{6^{n}(2^{k+1}R)^{n}}{(|x|+2^{k+1}R)^{n}}\leq \frac{C_{n}\ \log(e+|x|/R)}{(1+|x|/R)^{n}} \,. \tag{*}$$
In this bound, you can regard $n$ as a constant since a factor of $C_n$ is allowed on the RHS.
For $k$ such that $2^{k+1}R \le |x|+R$, replace the expression $(|x|+2^{k+1}R)$ in the denominator by $|x|+R$. Then all the resulting terms become
$$\frac{6^n (2R)^n}{|x|+R} \,,$$ and the number of such terms is at most $\log_2(e+|x|/R)$. This yields the RHS of $(*)$.
For $k$ such that $2^{k+1}R >|x|+R$, replace the expression $(|x|+2^{k+1}R)$ in the denominator by $2^{k+1}R$. Then these terms form a geometric series (with decreasing summands), that can be bounded by a constant multiple of the first term. That first term (for which $|x|+R <2^{k+1}R \le 2|x|+2|R|$) has the form $$\frac{C_n'}{(1+|x|/R)^n} $$ which is smaller than the RHS of $(*)$.