China Western Mathematical Olympiad

Denote $t=\lfloor \frac{n}{2}\rfloor +1$, and let $A_{n+j}=A_j$, $j=1,2,\dots ,n$. Noting that the distance between any vertex of a regular $n$-gon and a point on its side is not greater than its longest diagonal $d$, we therefore have, for any $1\le i\le n$, $$A_{i}P+P{B}_i=A_i{B}_i\le d.\stackrel{◯}{1}$$ Furthermore, using the fact that the sum of any two sides of a triangle is longer than the third side, we have, for any $1\le i\le n$, $$A_{i}P+P{A}_{i+1}\ge A_i{A}_{i+1}=d.\stackrel{◯}{2}$$ Summing up ①, ② for $i=1,2,\dots ,n$, we have $$\sum _{i=1}^n(A_{i}P+P{A}_{i+1})\ge nd\ge \sum _{i=1}^n(A_{i}P+P{B}_i),$$ i. e. $$2\sum _{i=1}^n{A}_i\ge \sum _{i=1}^n{A}_{i}P+\sum _{i=1}^{n}P{B}_i,$$ following which the proposition is proven.