How prove this$\frac{1}{P_{0}P_{1}}+\frac{1}{P_{0}P_{2}}+\cdots+\frac{1}{P_{0}P_{n}}<\sqrt{15n}$

Question

Let $P_{0},P_{1},P_{2},\cdots,P_{n}$ be $n+1$ points in the plane. Let $ d=1$ denote the minimal value of all the distances between any two points. Prove that

$$\dfrac{1}{P_{0}P_{1}}+\dfrac{1}{P_{0}P_{2}}+\cdots+\dfrac{1}{P_{0}P_{n}}<\sqrt{15n}$$

This problem background is from China high school math competition (Oct 14, 2012) problem 15,can see http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2822547&sid=bbbc81f99da00d657f61b4835931c87e#p2822547

also can see this two solution:http://wenku.baidu.com/view/82fb84d4240c844769eaeea3.html

But for my problem,I can't prove it.and I think this is nice problem,and Thank you for your help.

Answer 1

Up to relabelling, we can assume that $P_0 P_n$ is the greatest distance among $P_0 P_i$. We can also assume that $$\sum_{j=1}^{n-1}\frac{1}{P_0 P_j}<\sqrt{15(n-1)}$$ holds as induction hypothesis.

If $P_0 P_n \geq \frac{\sqrt{n}+\sqrt{n+1}}{\sqrt{15}}$ there is nothing to prove.

Otherwise, all the points belong to a circle $\Gamma$ centered in $P_0$ having radius $$R= \frac{\sqrt{n}+\sqrt{n+1}}{\sqrt{15}}<2\sqrt{\frac{n+1}{15}}$$ and area $[\Gamma]=\pi R^2$ less than $\frac{4\pi}{15}(n+1)$. Since at most $2\pi R$ points can lie in the annulus $A=\{x:d(x,P_0)\in[R-1/2,R]\}$, we have at least $(n+1-2\pi R)$ disjoint circles, having radius $\frac{1}{2}$, contained in $\Gamma$. By considering the area of the associated triangulation, we have: $$[\Gamma]= \pi R^2 > (n+1-2\pi R)\frac{\sqrt{3}}{2},$$ or: $$ \frac{2\pi}{\sqrt{3}}R^2 + 2\pi R > (n+1) $$ that is a contradiction once ... ahem ... $n>9881$. This leaves "a few" cases to be checked by hand.