<s>[INST] 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?
 [/INST] Assuming $a,b\gt0$, we get
$$
\begin{align}
\sum_{i=2^{n-1}+1}^{2^n}\left(\frac1{a+ib}-\frac{2^{-n+1}}{a+2b}\right)
&\ge\sum_{i=2^{n-1}+1}^{2^n}\left(\frac1{a+2^nb}-\frac{2^{-n+1}}{a+2b}\right)\\
&=\sum_{i=2^{n-1}+1}^{2^n}\frac{a(1-2^{-n+1})}{(a+2^nb)(a+2b)}\\
&=\frac{a(2^{n-1}-1)}{(a+2^nb)(a+2b)}\\[12pt]
&\ge0
\end{align}
$$
Therefore,
$$
\sum_{i=2^{n-1}+1}^{2^n}\frac1{a+ib}\ge\frac1{a+2b}
$$
</s>