The discussion revolves around a mathematical problem concerning real numbers \(x_1, ..., x_n\) that satisfy specific conditions: the sum of the numbers equals zero and the sum of their squares equals one. Participants are tasked with proving that for some indices \(k\) and \(l\), the product \(x_k x_l\) is less than or equal to \(-1/n\). The scope includes mathematical reasoning and proofs.
There is no consensus on the proof methods, as participants present different approaches and challenge each other's reasoning. Some participants agree on the necessity of proving the inequality, while others question specific logical steps.
Participants note that the problem's conditions lead to contradictions under certain assumptions, but the exact implications of these contradictions are debated. The discussion also highlights the complexity of the mathematical reasoning involved.
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
Sudharaka said:\[\left(\sum_{i=1}^{n}x_{i}\right)^2=\sum_{i=1}^{n}x_{i}^{2}+\sum_{k\neq l}x_{k}x_{l}\mbox{ where }k,l\in\{1,\cdots,n\}\]
\[\Rightarrow 0=1+\sum_{k\neq l}x_{k}x_{l}\]
\[\Rightarrow\sum_{k\neq l}x_{k}x_{l}=-1~-------(1)\]
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.
\[x_{k}x_{l}\leq -\frac{1}{n}\mbox{ for some }k,l\in\{1,\cdots,n\}\]
CaptainBlack said:In your example for \(n=2\) (which is essentially the only \(n=2\) possibility satisfying the given conditions):
\(i=1, j=2\), then \(x_ix_j=-\frac{1}{2}=-\frac{1}{n}\le -\frac{1}{n}\)
Which disposes of the case where \(n=2\)
CB
Sudharaka said:\[\left(\sum_{i=1}^{n}x_{i}\right)^2=\sum_{i=1}^{n}x_{i}^{2}+\sum_{k\neq l}x_{k}x_{l}\mbox{ where }k,l\in\{1,\cdots,n\}\]
Since \(\displaystyle\sum_{i=1}^{n}x_{i}=0\),
\[\sum_{i=1}^{n}x_{i}^{2}=-\sum_{k\neq l}x_{k}x_{l}~~~~~~~~~(1)\]
Suppose that, \(x_{k}x_{l}>-\frac{1}{n}~\forall~k,l\in\{1,\cdots,n\}\). Then,
\[\sum_{k\neq l}x_{k}x_{l}>-\frac{1}{n}\]
\[\Rightarrow -\sum_{k\neq l}x_{k}x_{l}<\frac{1}{n}~~~~~~~~~(2)\]
By (1) and (2);
\[\sum_{i=1}^{n}x_{i}^{2}<\frac{1}{n}<1\]
This is a contradiction. Hence,
\[x_{k}x_{l}\leq -\frac{1}{n}\mbox{ for some }k,l\in\{1,\cdots,n\}\]
CaptainBlack said:Hint:
Let \(a=\max(x_i, i=1, .. , n) \) and \( b=\min(x_i, i=1, .. , n) \) and then consider \( f(x)=(x-a)(x-b) \)
CB
Sudharaka said:\[f(x_{i})=(x_{i}-a)(x_{i}-b)=x_{i}^{2}-x_{i}(a+b)+ab\]\[\Rightarrow\sum_{i=1}^{n}f(x_{i})=\sum_{i=1}^{n}x_{i}^{2}-(a+b)\sum_{i=1}^{n}x_{i}+ab\sum_{i=1}^{n}1\]
\[\Rightarrow\sum_{i=1}^{n}f(x_{i})=1+abn~~~~~~~(1)\]
Also note that since \(a=\max(x_i, i=1,\cdots, n)\) and \( b=\min(x_i, i=1,\cdots, n)\),
\[x_{i}-a\leq 0\mbox{ and }x_{i}-b\geq 0~\forall~i=1,\cdots,n\]
\[\therefore\sum_{i=1}^{n}f(x_{i})=\sum_{i=1}^{n}(x_{i}-a)(x_{i}-b)\leq 0~~~~~~~(2)\]
By (1) and (2);
\[1+abn\leq 0\]
\[\Rightarrow ab\leq -\frac{1}{n}\]
So we have found a \(x_{k}\mbox{ and a }x_{l}\) (equal to \(a\) and \(b\)) so that \(x_{k}x_{l}\leq -\frac{1}{n}\)
I am pretty sure that this is the answer that you have in mind. Isn't? :)
Sudharaka said:\[\left(\sum_{i=1}^{n}x_{i}\right)^2=\sum_{i=1}^{n}x_{i}^{2}+\sum_{k\neq l}x_{k}x_{l}\mbox{ where }k,l\in\{1,\cdots,n\}\]
Since \(\displaystyle\sum_{i=1}^{n}x_{i}=0\),
\[\sum_{i=1}^{n}x_{i}^{2}=-\sum_{k\neq l}x_{k}x_{l}~~~~~~~~~(1)\]
Suppose that, \(x_{k}x_{l}>-\frac{1}{n}~\forall~k,l\in\{1,\cdots,n\}\). Then,
\[\sum_{k\neq l}x_{k}x_{l}>-\frac{1}{n}\]
\[\Rightarrow -\sum_{k\neq l}x_{k}x_{l}<\frac{1}{n}~~~~~~~~~(2)\]
By (1) and (2);
\[\sum_{i=1}^{n}x_{i}^{2}<\frac{1}{n}<1\]
This is a contradiction. Hence,
\[x_{k}x_{l}\leq -\frac{1}{n}\mbox{ for some }k,l\in\{1,\cdots,n\}\]
caffeinemachine said:Hi Sudharaka.
I couldn't get the "then" part in the following:
Suppose that $x_k x_l > - \frac{1}{n} \forall k,l \in \{ 1, \ldots, n \}$. THEN $\displaystyle\sum_{k \neq l}x_k x_l > - \frac{1}{n}$
Sudharaka said:Hi! (Wave)
Since $x_{k}x_{l}>- \frac{1}{n}~\forall~k,l \in \{ 1, \ldots, n \}$, the addition of any finite number of terms in the set $\{x_k x_l:k,l\in\{1,\ldots,n\}\}$ is also greater than $-\frac{1}{n}$. This stems from the result; if $a>b\mbox{ and }c>b\Rightarrow a+c>b\mbox{ where }a,b,c\in\Re$.
caffeinemachine said:I don't think that's right. $-0.7>-1, -0.8>-1$ but $-(0.7+0.8) \not > -1$
Sudharaka said:Ahhh...(Headbang) Indeed it is incorrect. It should be $a>b\mbox{ and }c>b\Rightarrow a+c>2b\mbox{ where }a,b,c\in\Re$ I don't how I made such a terrible mistake but thanks for pointing that out. That being said the solution is incorrect with reference to this mistake. I think CB's constructive proof is the only method to tackle this problem.