CaptainBlack said:
Let \(x_1, ..., x_n\) be real numbers such that:
\(\sum_{i=1}^n x_i=0\) and \(\sum_{i=1}^n x_i^2=1\)
Prove that for some \( k,l \) both in \( \{1, .. , n\} \) that \(x_k x_l\le -1/n\)
CB
\[\left(\sum_{i=1}^{n}x_{i}\right)^2=\sum_{i=1}^{n}x_{i}^{2}+\sum_{k\neq l}x_{k}x_{l}\mbox{ where }k,l\in\{1,\cdots,n\}\]
\[\Rightarrow 0=1+\sum_{k\neq l}x_{k}x_{l}\]
\[\Rightarrow\sum_{k\neq l}x_{k}x_{l}=-1~-------(1)\]
Let \(A=\{x_{k}x_{l}:k\neq l\mbox{ and }k,l\in\{1,\cdots,n\}\}\). Notice that, \(|A|=n(n-1)\). Take \(m=\mbox{min}(A)\). Then,
\[m\leq x_{k}x_{l}\forall~k,l\in\{1,\cdots,n\}\mbox{ where }k\neq l\]
Since \(|A|=n(n-1)\)
\[mn(n-1)\leq\sum_{k\neq l}x_{k}x_{l}~-------(2)\]
By (1) and (2);
\[mn(n-1)\leq -1\]
\[m\leq\frac{-1}{n(n-1)}~------(3)\]
Suppose that, \(x_{k}x_{l}>-\frac{1}{n}~\forall~k,l\in\{1,\cdots,n\}\). Then,
\[m>-\frac{1}{n}~------(4)\]
By (3) and (4);
\[-\frac{1}{n}<m\leq\frac{-1}{n(n-1)}\]
\[\Rightarrow n>2\]
Therefore in order for our assumption to be true \(n\) should be greater than 2. Consider the set, \(\left\{\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right\}\). This set has two elements but, \(\displaystyle\sum_{i=1}^{2}x_{i}=0\mbox{ and }\sum_{i=1}^{n}x_{i}^{2}=1\). Therefore our assumption is wrong.
\[x_{k}x_{l}\leq -\frac{1}{n}\mbox{ for some }k,l\in\{1,\cdots,n\}\]