We first show that the angles of $ O_1O_2 \cdots O_n $ are identical to those of $A_1A_2 \cdots A_n $. Then we show that this angle condition translated to an ordering of the lengths of the cirumradii of the respective triangles $A_{i-1}A_iA_{i+1} $ that we can easily derive a contradiction from.

Begin with noticing that $O_{i-1}O_{i} $ is the perpendicular bisector of $A_{i-1}A_i $. Analogously for $O_{i-2}O_{i-1} $. It easily follows from here that $\angle O_{i-2}O_{i-1}O_{i} = \angle A_{i-2}A_{i-1}A_{i} \text{ or } 180^{\circ} - \angle A_{i-2}A_{i-1}A_{i} $. This is where we use the conditions of the angles of $ A_1A_2 \cdots A_n $ being obtuse; indeed, we are able to get as a result that $ \angle O_{i-2}O_{i-1}O_{i} \leq \angle A_{i-2}A_{i-1}A_{i} $ (*). Summing (*) over all indices gives that equality holds for all indices. Let the circumradius of $ A_{i-1}A_{i}A_{i+1} = R_i $. It is easy to conclude now that there are 2 possible cases: $ R_{i+1} > R_i > R_{i-1}$ or $R_{i-1} > R_i > R_{i+1}$. Now, consider taking $ i $ such that $R_i = \max_{j} R_j $, which gives a contradiction in the previous 2 cases by maximality.