Prove that $\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$ 
Prove that $$\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} \ge \frac{1}{a + 2b}$$

I tried to to prove the above statement using the AM-HM inequality:
$$\begin{align}\frac{1}{2^n - 2^{n-1}}\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \frac{2^n - 2^{n-1}}{\sum_{i = 2^{n-1} + 1}^{2^n}(a + ib)}\\
\sum_{i = 2^{n-1} + 1}^{2^n}\frac{1}{a + ib} &\ge \frac{(2^n - 2^{n-1})^2}{\frac{2^n -2^{n-1}}{2}(2a + (2^n + 2^{n-1} + 1)b)}\\
&=\frac{2^{n+1} - 2^n}{2a + (2^n + 2^{n-1} + 1)b}\end{align}$$
after which I am more or less stuck. How can I continue on from here, or is there another method?

A generalized version of the result might be easier to prove:

For every positive integers $k$ and $m$, $$\sum_{i=k+1}^{k(m+1)}\frac1{a+ib}\geqslant\frac{m}{a+(m+1)b}.$$

The question asks about the case $k=2^{n-1}$ and $m=1$.
To prove the claim, note that $a+ib\leqslant a+k(m+1)b$ for every $i$ used in the sum $S$ of the LHS and that the sum $S$ has $km$ terms hence $$S\geqslant\frac{km}{a+k(m+1)b}=\frac{m}{(m+1)b}\,\left(1-\frac{a}{a+k(m+1)b}\right).$$ The RHS is an increasing function of $k$ hence it is at least equal to its value when $k=\color{red}{\bf1}$, that is, $$S\geqslant\frac{\color{red}{\bf1}\cdot m}{a+\color{red}{\bf1}\cdot (m+1)\,b}=\frac{m}{a+(m+1)b}.$$
