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I have to show

$(2 \cdot n) \cdot 2 \cdot \sum_{i=1}^{n}t_i^2-\left ( 2 \cdot \sum_{i=1}^{n} t_i \right )^2>0$

Which is probably not advanced but I am extremly new to working with sums in linear algebra. This is regarding a problem showing that a point is a minimum in a function using ABC-criteria. I need to show the above property for the last step. I am not experienced AT ALL with this and would really love some help.

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  • $\begingroup$ Just apply Cauchy -Schwarz inequality for $\sum (1) (t_i)$ $\endgroup$ – Kavi Rama Murthy Nov 28 '20 at 0:29
  • $\begingroup$ Even simpler: do you know what $1 + 2 + 3 + \cdots + n$ equals? How about $1^2 + 2^2 + 3^2 + \cdots + n^2$? $\endgroup$ – Toby Mak Nov 28 '20 at 0:29
  • $\begingroup$ @Toby Mak n(n+1) / 2 and (n(n+1)(2n+1))/6 $\endgroup$ – user831870 Nov 28 '20 at 0:34
  • $\begingroup$ @KaviRamaMurthy How would one do that? I am really not familiar with this line $\endgroup$ – user831870 Nov 28 '20 at 0:36
  • $\begingroup$ Actually it's not that simple, but eventually after expanding you get $\frac{1}{3}n^2(n^2-1)$, which is $> 0$ if $n > 1$. $\endgroup$ – Toby Mak Nov 28 '20 at 0:36
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This is just Cauchy Schwarz inequality, $ \sum a_i^2 \sum b_i^2 \geq (\sum a_i b_i)^2$

Set $ a_i = 2, b_i = t_i$, and we get that

$$ 4n \sum t_i^2 \geq (\sum 2 t_i)^2. $$

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  • $\begingroup$ Ohhh I see. The $n$ was theowing me off on how to apply it! $\endgroup$ – user831870 Nov 28 '20 at 14:23
  • $\begingroup$ Well but by this it can be 0 then? $\endgroup$ – user831870 Nov 28 '20 at 19:44
  • $\begingroup$ @bymathformath What do you mean by "but by this it can be 0 then"? If you're asking if equality can hold, then yes equality can hold. $\endgroup$ – Calvin Lin Nov 29 '20 at 10:56

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