We claim the minimum value is $5$ which can be achieved by setting $b_{i} = \frac{1}{4}$ where $2 \leq i \leq m$ Start by seeing that $$b_m = a_{m}a_{m-1}\cdots a_{2} - \frac{1}{4}\left(a_{m}a_{m-1}\cdots a_{3} + \cdots + a_{m}a_{m-1} + a_m + 1\right)$$$$ = a_{m}a_{m-1}\cdots a_{2} - \frac{a_{m}a_{m-1}\cdots a_{1}}{4}\left(\sum_{k=1}^{m} \frac{1}{a_{1}a_{2}\cdots a_{k}}\right) \implies \sum_{k=1}^{m} \frac{1}{a_{1}a_{2}\cdots a_{k}} = 5 - \frac{4b_m}{a_{m}a_{m-1} \cdots a_2}$$ Now note that $$a_{2}b_{1}a_{3}b_{2}\cdots a_{m}b_{m-1} = \prod_{i=2}^{k} \left(b_i + \frac{1}{4}\right) \implies \frac{b_m}{a_{1}a_{2}\cdots a_{m}} = \frac{1}{\prod_{i=1}^{m} \left(1 + \frac{1}{4b_{i}}\right)}$$. Now the claim to finish off , $$\sqrt{b_{1}\cdots b_{m}} \geq \frac{1}{\prod_{i=1}^{m} \left(1 + \frac{1}{4b_{i}}\right)}$$since $$1 + \frac{1}{4b_i} \geq \frac{1}{\sqrt{b_i}}$$. Therefore $$4\sqrt{b_1b_2\cdots b_m}+\sum_{k=1}^m\frac{1}{a_1a_2\cdots a_k} = 4\left(\sqrt{b_{1} \cdots b_{m}} - \frac{1}{\prod_{i=1}^{m} \left(1 + \frac{1}{4b_{i}}\right)} + 1\right) + 1 \geq 5$$. Edit : Fixed @below , Thanks I missed a tiny detail in a rush.

wrong,min is 5