The most crucial idea is the following lemma: Lemma: Let $ n$ be some positive integer, greater than $ 3$.For an arbitrary set of points, $ A_1,A_2,\dots, A_n$, define point $ P$, which lies in the plane of points $ A_1,A_2,\dots, A_n$, as a point, sum of distances from which to the points $ A_i$ reaches minimum. Then the following equation holds: \[ \sum_{i = 1}^{n}\frac {\overline{PA_i}}{|PA_i|} = 0 \] Proof: Let me just to sketch it out: Suppose that $ (x,y)$ are the coordinates of the point $ P$, and $ (x_i,y_i)$, where $ i = 0,1,2,\dots,n$, are the coordinates of points $ Q,A_1,A_2,\dots, A_n$, respectively. Denote by $ f_Q(x,y) = |\overline{PQ}|$. If positive reals $ \alpha, \beta$ satisfy to the following conditions: 1. $ \alpha + \beta = 1$ 2. There exist two points $ P'(x',y')$ and $ P''(x'',y'')$, so that: \[ x = \alpha\cdot x' + \beta\cdot x'', y = \alpha\cdot y' + \beta\cdot y'' \] Therefore, $ f_Q(\alpha\cdot x' + \beta\cdot x'', \alpha\cdot y' + \beta\cdot y'') = |\overline{PQ}| < \alpha|\overline{QP'}| + \beta|\overline{QP''}| =$ $ = \alpha f_Q\left(x',y'\right) + \beta f_Q\left(x'',y''\right)$. It means that $ f_Q$ is a strictly convex function, hence, it has only one minimum point. Afterwards, we might use the fact that, in order to reach minimum at some point, for every unit vector $ \overline{e} = (\cos{\alpha},\sin{\alpha})$, derivative of the function in the line of $ e$ must be nonnegative. Using this property and recalling that sum of derivatives is derivative of the sum, we get the desired equation. $ \square$. Obviously if these points, $ A_1,A_2,\dots A_n$ form regular $ n$-gon, then the point $ X$ coincides with center of $ n$-gon. I am going to prove a slightly stronger result, namely, define points $ B_i$ as intersection with circumcircle of $ n$-gon, instead of intersection with its boundary. If $ O$ is the center and $ R$ is the circumradius, then $ \sum_{i = 1}^{n}PA_i + \sum_{i = 1}^{n}PB_i\leq 2 \sum_{i = 1}^{n} OA_i\leq 2\sum_{i = 1}^{n} PA_i$. Therefore, $ \sum_{i = 1}^{n} PA_i\ge \sum_{i = 1}^{n} PB_i$ $ \blacksquare$

Is there any easier, non-calculus proof of Erken's lemma?

We look $A_i=A_j$ iff $n \mid i-j$ Note that if $k=\lfloor \frac{n}{2} \rfloor +1$, $A_1A_k$ is longest diagonal of that polygons We have $A_iP+PB_i =A_iB_i \leq d$ and $A_iP+PA_{i+k}\geq A_iA_{i+k}=d$ for each $i=1,2,...,n$ Summing up the inequality give us $2\sum_{i=1}^{n}{PA_i} \geq nd \geq \sum_{i=1}^{n}{PA_i +PB_i}$ So $\sum_{i=1}^n{PA_i} \geq \sum_{i=1}^{n}{PB_i}$

ThE-dArK-lOrD wrote: Note that if $k=\lfloor \frac{n}{2} \rfloor +1$, $A_1A_k$ is longest diagonal of that polygons We have $A_iP+PB_i =A_iB_i \leq d$ and $A_iP+PA_{i+k}\geq A_iA_{i+k}=d$ for each $i=1,2,...,n$ when $n$ is even, $A_iA_{i+k}$ is not the longest diagonal of that polygon, but $A_iA_{i+k-1} is$.