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I was messing around today and came across the following which I believe to be true: $$\left(\sum x_i\right)^2-2\sum_{cyc}x_ix_j=\sum x_i^2$$ for a set of $x=(x_1,x_2,...x_n)$ where $n\ge2$ and it is easy to prove for small values manually but how would I prove this rigourously? I thought about using the usual method of a base case, $n=k$ then $n=k+1$ but I am not quite sure how to do this?


$$n=2$$ $$\sum x_i=x+y,\sum_{cyc}x_ix_j=xy,\sum x_i^2=x^2+y^2$$ now: $$(x+y)^2-2xy=x^2+y^2$$ is easily shown. I have done the same for $n=3$ but how should I go about doing the general case?

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  • $\begingroup$ The sum $\sum x_ix_j$ has to be symmetric, otherwise it is not true, e.g $n=4$ fails. $\endgroup$ – Arben Ajredini May 30 '20 at 18:46
  • $\begingroup$ @ArbenAjredini ah I see, what determines whether or not the sum is symmetricl? $\endgroup$ – Henry Lee May 30 '20 at 18:47
  • $\begingroup$ The symmetric sum for $n=4$ would be $x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4$ and the cyclic sum is $x_1x_2+x_2x_3+x_3x_4+x_4x_1$ . $\endgroup$ – Arben Ajredini May 30 '20 at 18:52
  • $\begingroup$ Ah so is there a different term I can use to describe the sum of all pairs where $i\ne j$? $\endgroup$ – Henry Lee May 30 '20 at 19:39
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    $\begingroup$ You can use $\sum_{sym} x_ix_j$ $\endgroup$ – Arben Ajredini May 31 '20 at 14:36
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If you expand $$(x_1 + \cdots + x_n)(x_1 + \cdots + x_n)$$ you get $n^2$ addends: $x_1^2, \ldots, x_n^2$, as well as two each of $x_i x_j$ for $i \ne j$. The cyclic sum accounts for some of these latter terms, but misses terms like $x_1 x_3$ or $x_2 x_4$.

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