armandtrepy567
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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?
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?