Summation proof

**duchops** · February 10th 2010, 10:58 AM

Prove that $n*\sum_{i=0}^nx_i^2 \geq (\sum_{i=0}^nx_i)^2$

I tried proving it directly using summation identities. I then tired using induction, but I kept getting stuck.

A proof similar to this may be in the forum, but it is difficult to search for. Thanks.

-Chops

**icemanfan** · February 10th 2010, 11:52 AM

You cannot prove this, as it is false. However, you can prove

$n*\sum_{i=1}^nx_i^2 \geq (\sum_{i=1}^nx_i)^2$

by using the following fact repeatedly:

$(x_2 - x_1)^2 \geq 0$

$x_1^2 - 2 x_1 x_2 + x_2^2 \geq 0$

$x_1^2 + x_2^2 \geq 2x_1 x_2$

**Archie Meade** · February 10th 2010, 12:50 PM

Originally Posted by duchops

Prove that $n*\sum_{i=0}^nx_i^2 \geq (\sum_{i=0}^nx_i)^2$

I tried proving it directly using summation identities. I then tired using induction, but I kept getting stuck.

A proof similar to this may be in the forum, but it is difficult to search for. Thanks.

-Chops

$\sum_{i=0}^nx_i=\frac{n(n+1)}{2}\ and\ \sum_{i=0}^n{x_i}^2=\frac{n(n+1)(2n+1)}{6}$

$n\sum_{i=0}^nx_i^2=\frac{n^2(n+1)(2n+1)}{6}\$

$\left(\sum_{i=0}^nx_i\right)^2=\frac{n^2(n+1)(n+1) }{4}$

$\frac{n^2(n+1)(2n+1)}{6}\ \ge\ \frac{n^2(n+1)(n+1)}{4}$ ?

$\frac{n^2(n+1)}{2}\ is\ common$

$\frac{2n+1}{3}\ \ge\ \frac{n+1}{2}$ ?

$\frac{4n+2}{6}\ \ge\ \frac{3n+3}{6}$ ?

$4n+2\ \ge\ 3n+3$ ?

$n\ \ge\ 1$ ?

The inequality is true for all natural numbers n from 1 up, and for all natural numbers i. It does not matter if you sum from i=0 or from i=1, since the first term in both sums is zero.

**duchops** · February 10th 2010, 12:58 PM

Ah, sorry. Had the limits of the sum incorrect. Thanks for the tip.

**icemanfan** · February 10th 2010, 01:01 PM

Archie, your analysis is incorrect. Check the sums in the problem again. We are not given the values of the $x_i$ .

**Archie Meade** · February 10th 2010, 01:09 PM

Yes, it assumes the terms are consequtive natural numbers from 0 to n,
nothing more,
if not, another way is necessary.

**duchops** · February 10th 2010, 01:14 PM

No, it says $\sum_{i=1}^nx_i$ not $\sum_{i=1}^ni$

Icemanfan is correct.

**Archie Meade** · February 10th 2010, 01:21 PM

Yep,
what i wrote takes the terms as a consequtive series of natural numbers from 0 to n. The question is asking to prove it for n arbitrary terms then ?
ok, I wrongly assumed it was your way of writing the sequence of natural numbers. I'll take another shot in a while.

**Archie Meade** · February 10th 2010, 01:57 PM

$\sum_{i=0}^nx_i^2=x_0^2+x_1^2+x_2^2+....+x_n^2$

$\left(\sum_{i=0}^nx_i\right)^2=\left(x_0+x_1+x_2+. ...+x_n\right)\left(x_0+x_1+x_2+....+x_n\right)$

$(a+b)(a+b)=a^2+b^2+2ab$

$2(a^2+b^2)=a^2+b^2+a^2+b^2$

$a^2+b^2\ is\ common$

$Is\ a^2+b^2\ \ge\ 2ab\ ?\ Yes,\ since\ (a-b)^2=a^2+b^2-2ab\ \ge\ 0$

$(a+b+c)^2=(a+b+c)(a+b+c)=a^2+ab+ac++ba+b^2+bc+ca+c b+c^2$ $=a^2+b^2+c^2+2ab+2ac+2bc$

$3(a^2+b^2)=a^2+b^2+c^2+[a^2+b^2]+[a^2+c^2]+[b^2+c^2]$

$Is\ a^2+b^2\ \ge\ 2ab\ ?\ Yes$
$Is\ a^2+c^2\ \ge\ 2ac\ ?\ Yes$
$Is\ b^2+c^2\ \ge\ 2ac\ ?\ Yes$

$(a+b+c+d)^2=(a+b+c+d)(a+c+b+d)$
$=a^2+ab+ac+ad+ba+b^2+bc+bd+ca+cb+c^2+cd+da+db+dc+d ^2$
$=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd$

$4(a^2+b^2+c^2+d^2)=a^2+b^2+c^2+d^2+(a^2+b^2)+(a^2+ c^2)+$ $(a^2+d^2)+(b^2+c^2)+(b^2+d^2)+(c^2+d^2)$

$a^2+b^2\ \ge\ 2ab$
$a^2+c^2\ \ge\ 2ac$
$a^2+d^2\ \ge\ 2ad$
$b^2+c^2\ \ge\ 2bc$
$b^2+d^2\ \ge\ 2bd$
$c^2+d^2\ \ge\ 2cd$

This pattern is consistent.

**Archie Meade** · February 10th 2010, 02:08 PM

Wait, that's not finished!

**Archie Meade** · February 11th 2010, 05:38 AM

Hi duchops,
there is a general pattern to the inequality.
The above post is another way to view it.

Actually, that is the same thing icemanfan said,
so yes, he's quite right.

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