$\textbf{Proof :}$ Let $R$ be radius of circle $(A_1A_2\ldots A_n)$. Let $r_i$ be radius of circle $(B_{i-1}A_iB_{i})$. We know that $\angle A_{i-1}A_iA_{i+1} = \angle B_{i-1}A_iB_{i}$, so $\frac{r_i}{R} = \frac{B_{i-1}B_{i}}{A_{i-1}A_{i+1}}$. So to prove that $\frac{B_1B_2}{A_1A_3} + \frac{B_2B_3}{A_2A_4} + ... + \frac{B_nB_1}{A_nA_2} > 1$ we just need to prove that $r_1+r_2+\ldots + r_n> R$. Note that $B_iA_{i+1}+A_{i+1}B_{i+1}\leq 4r_i$, so it's enough to prove that $\sum_i A_iA_{i+1}>4R$. Also well known that if circumcenter is strictly in interior of $A_1A_2 ...A_n$, than perimeter of $A_1A_2 ...A_n$ is greater than $4R$. $\Box$

Let $r_i$ denote the radius of $(B_{i-1}A_iB_i)$ and $R$ be the radius of $(A_1A_2\dots A_n)$. Observe that $$\frac{B_iB_{i+1}}{A_iA_{i+2}}=\frac{2r_i \sin A_{i+1}}{2R \sin A_{i+1}}=\frac{r_i}{R},$$hence it suffices to show $r_1+\dots+r_n >R$. However, by triangle inequality, we have $r_i+r_{i+1}>\frac{1}{2}A_iA_{i+1}$, hence it suffices to show $(A_1A_2 \dots A_n)$ has perimeter at least $4R$. By applying triangle inequality successively, we see that it is enough to show this for $n=3$. Let $ABC$ be a triangle with all angles acute, say $a, b, c$. Note that $$AB+BC+CA =2R(\sin a+\sin b+\sin c),$$hence, it suffices to minimise $\sin a+\sin b+\sin c$ subject to $0 \le a\le b \le c \le 90^{\circ}$ and $a+b+c=180^{\circ}$. Notice that $$\sin (c+d)-\sin c \le \sin a-\sin (a-d) \iff 2\sin \left(\frac{1}{2}d\right) \cos \left(c+\frac{d}{2}\right) \le 2\sin \left(\frac{1}{2}d\right) \cos \left(a-\frac{1}{2}d\right),$$which is true as $\cos$ is a decreasing function on $[0, \tfrac{\pi}{2}]$, for all $0 \le d \le \min(a, 90^{\circ}-c)$. Hence, we can make $b=c=90^{\circ}$ and $a=0^{\circ}$ without increasing the sum, so the minimum is at least $2$, exactly as per our requirement. $\blacksquare$