Proving $\frac{B_1B_2}{A_1A_3}+ \frac{B_2B_3}{A_2A_4}+...+\frac{B_nB_1}{A_nA_2}>1$ Let $A_1A_2 . . . A_n$ be a cyclic convex polygon whose circumcenter is strictly in its interior. Let $B_1, B_2, ..., B_n$ be arbitrary points on the sides $A_1A_2, A_2A_3, ..., A_nA_1$,  respectively, other than the vertices (that is With $B_i\neq A_k$).

Prove that
  $$\frac{B_1B_2}{A_1A_3}+ \frac{B_2B_3}{A_2A_4}+...+\frac{B_nB_1}{A_nA_2}>1$$

Any idea or Hint?
Note the might be similar to this with the difference here we are in a more complicated setting.
 A: There’s probably a better solution but here we go.
Note that by Law of Sines, the inequality is equivalent to $$\frac{1}{R}\sum_{i=1}^{n}r_{i}>1$$, where $R,r_{i}$ denote circumradii of $A_{1}...A_{n}$ and $B_{i}A_{i}B_{i+1}$.
If $n=3$, the triangle is acute and circumcircles of $B_{1}A_{1}B_{2}, B_{2}A_{2}B_{3}, B_{3}A_{3}B_{1}$  concurs at a point $P$ (simple angle chasing). $A_{1}P, A_{2}P, A_{3}P$ are chords in their respective circles, so their lengths are less than or equal to $2r_{1}, 2r_{2}, 2r_{3}$.
It remains to show that $PA_{1}+PA_{2}+PA_{3}>2R$. It’s obvious that there is a side, say $BC$, such that $OP$, where $O$ is the triangle’s circumcenter, does not pass internally. Because the triangle is acute $A’$, the antipodal point of $A$, lies on circular segment $BC$ not containing $A$. WLOG, let $PB\geq PC$. Then a circle $P$ with radius $PB$ contains every point on that circular segment including $A’$. Hence, $2R=AA’\leq PA+PA’\leq PA+PB<PA+PB+PC$ as desired. 
(The case $O=P$ is trivial.)
Consider when $n\geq 4$. There clearly exists a quadrilateral $A_{i}A_{j}A_{k}A_{l}, i<j<k<l$, with the circumcenter inside. Supporting angles at the center of $A_iA_j, A_jA_k, A_kA_l, A_lA_i$ sum up to $2\pi$ so there is a pair of opposite sides whose supporting angles sum up to at least $\pi$. WLOG, we therefore have $A_iA_j+A_kA_l\geq 2R$. Considering chords of circumcircles, we have 
$$A_iA_j\leq A_iB_i+B_iB_{i+1}+...+B_{j-1}A_j\leq 2r_{i-1}+...+2r_{j-1}.$$
Equality occurs if and only if $j=i+1$ and $\angle A_iB_{i-1}B_i=\angle B_iB_{i+1}A_{i+1}=\pi /2$.
Similar result for $A_kA_l$ can be drawn, thus
$$2R\leq 2r_{i-1}+...+2r_{j-1}+2r_{k-1}+...+2r_{l-1}\leq 2\sum_{i=1}^{n}r_{i}.$$
Equality can only happen with $n=4$. Then we clearly see that angle requirements lead to a contradiction. The conclusion follows.
