1. ## Summation proof

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

2. 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:

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

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

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

3. 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.

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

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

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

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

Icemanfan is correct.

8. 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.

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

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

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

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

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

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

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

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

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

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

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

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

This pattern is consistent.

10. Wait, that's not finished!

11. 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.