Define $$a_{2n}=\sqrt{a_nA_n}, \; \;\; A_{2n}=\frac{2A_na_{2n}}{A_n+a_{2n}},$$with $a_4=2$ and $A_4=4$. Prove that $a_4,a_8,a_{16},a_{32},\dots$ is increasing.
My Attempt
We write $$\{a_{2^{n+1}}\}=\{a_4,a_8,\dots\}.$$ Then the sequence is increasing iff $$a_{2^{n+1}}\geqslant a_{2^{n}}.$$ Using the definitions, I get to the point where proving that the sequence is increasing is reduced to showing $$A_{2^{n-1}}\geqslant a_{2^n},$$ but I am stuck on how to show this. Any help will be appreciated.
Defining sequences $b_n=\ln a_{2^{n+1}}$ and $B_n=\ln A_{2^{n+1}}$, the problem becomes showing that $(b_n)$ is increasing. Since $b_{n+1}=\frac{1}{2}(b_n+B_n)$, it is equivalent to show that $b_n < B_n$ for all $n \in \mathbb{Z}^+$. This is true for $n=1$, so we proceed by induction. Since $b_{n+1}=\frac{1}{2}(b_n+B_n)$ and $B_{n+1}=\ln 2 + B_n + b_{n+1} - \ln(e^{b_n}+e^{B_n})$, it is equivalent to show that $\ln(e^{b_n}+e^{B_n}) < \ln2 + B_n$, which is plainly true by the assumption that $b_n < B_n$. QED