Given n+1 points, bound the product of the distances from one of them

[armandtrepy567]

We have $\displaystyle n+1$ real numbers $\displaystyle x_1,\cdots,x_{n+1}$ such that $\displaystyle -1\leq x_i\leq 1$ for all $\displaystyle 1\leq i\leq n+1$.

Prove: There exists some $\displaystyle j$ such that $\displaystyle \prod_{\substack{i=1\\i\neq j}}^{n+1}\left |x_j-x_i\right |\leq \frac{n+1}{2^{n-1}}$.

Some base steps are easy. Assume that this holds for $\displaystyle n-1$.

Applying this inductive hypothesis to $\displaystyle \left\{x_i:i\in\mathbb{N}_{\leq n+1}\right\}\setminus\left\{x_j\right\}$ for each $\displaystyle j\in\mathbb{N}_{\leq n+1}$, we conclude that for each $\displaystyle j\in\mathbb{N}_{\leq n+1}$ there exists $\displaystyle h\left(j\right)\in\mathbb{N}_{\leq n+1}\setminus\left\{j\right\}$ such that $\displaystyle \prod_{\substack{i=1\\i\neq j\\i\neq h\left(j\right)}}^{n+1}\left |x_{h\left(j\right)}-x_i\right |\leq\frac{n}{2^{n-2}}$.

Now, if for some $\displaystyle j\in\mathbb{N}_{\leq n+1}$ we have $\displaystyle \left |x_{h\left(j\right)}-x_j\right |\leq \frac{1}{2}$, then $\displaystyle \prod_{\substack{i=1\\i\neq h\left(j\right)}}^{n+1}\left |x_{h\left(j\right)}-x_i\right |\leq \frac{1}{2}\cdot\frac{n}{2^{n-2}}\leq \frac{1+\frac{1}{n}}{2}\frac{n}{2^{n-2}}=\frac{n+1}{2^{n-1}}$, which gives what we want. But what about the case in which $\displaystyle \left |x_{h\left(j\right)}-x_j\right |>\frac{1}{2}$ for all $\displaystyle j\in\mathbb{N}_{\leq n+1}$?

Is there any other way?