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?

For $a,b,x>0$, the function $f(x)=\frac{1}{a+bx}$ is convex. So by Jensen's inequality
$$
\frac{1}{2^{n-1}}\sum_{i=2^{n-1}+1}^{2^n}f(i)\geq f\left(\frac{1}{2^{n-1}}\sum_{i=2^{n-1}+1}^{2^n}i\right)=f(0.5+3\times2^{n-2})=\frac{1}{a+(0.5+3\times2^{n-2})b}.
$$
It remains to show $2^{n-1}$ times the rightmost expression above is greater than or equal to the RHS of the desired inequality. This amounts to computing:
$$
2^{n-1}(a+2b)-(a+(0.5+3\times2^{n-2})b)=a(2^{n-1}-1)+b(2^n-0.5-3\times2^{n-2})\\
=a(2^{n-1}-1)+b(2^{n-2}-0.5)=(2^{n-1}-1)(a+b/2)\geq 0.
$$
