# Inequality of Sums (Cauchy-Schwarz?)

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.

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

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
$$4n \sum t_i^2 \geq (\sum 2 t_i)^2.$$
• Ohhh I see. The $n$ was theowing me off on how to apply it! – user831870 Nov 28 '20 at 14:23