# Derive a relation using Cauchy Schwarz inequality

• Oct 23rd 2013, 07:55 AM
MichaelH
Derive a relation using Cauchy Schwarz inequality
So I am trying to derive the following relation using Cauchy Schwarz inequality for any collection of N real numbers:

$\displaystyle \left(\frac{a_1+a_2+\cdots+a_N}{N}\right)^2 =< \frac{a^2_1+a^2_2+\cdots+a^2_N}{N}.$

This says that the square of the average is less or equal than the average of the squares.

So I want to use $\displaystyle |v \cdot w|^2 =(v \cdot v)*(w \cdot w)$ to derive this.

So $\displaystyle v \cdot w$ needs to be $\displaystyle \frac{a_1+a_2+\cdots+a_N}{N}$ therefore, I have $\displaystyle v \cdot v$ as $\displaystyle a^2_1+a^2_2+\cdots+a^2_N$ and $\displaystyle w \cdot w$ as $\displaystyle \frac{1}{N^2}$ providing that:

$\displaystyle v: a_1+a_2+\cdots+a_N$

$\displaystyle w: \frac{1}{N}\\frac$
$\displaystyle {1}{N}$

Am I going in the right direction? Can anyone provide me some guidance?
• Oct 23rd 2013, 08:20 AM
SlipEternal
Re: Derive a relation using Cauchy Schwarz inequality
Quote:

Originally Posted by MichaelH
So I am trying to derive the following relation using Cauchy Schwarz inequality for any collection of N real numbers:

$\displaystyle \left(\frac{a_1+a_2+\cdots+a_N}{N}\right)^2 =< \frac{a^2_1+a^2_2+\cdots+a^2_N}{N}.$

This says that the square of the average is less or equal than the average of the squares.

So I want to use $\displaystyle |v \cdot w|^2 =(v \cdot v)*(w \cdot w)$ to derive this.

So $\displaystyle v \cdot w$ needs to be $\displaystyle \frac{a_1+a_2+\cdots+a_N}{N}$ therefore, I have $\displaystyle v \cdot v$ as $\displaystyle a^2_1+a^2_2+\cdots+a^2_N$ and $\displaystyle w \cdot w$ as $\displaystyle \frac{1}{N^2}$ providing that:

$\displaystyle v: a_1+a_2+\cdots+a_N$

$\displaystyle w: \frac{1}{N}$

Am I going in the right direction? Can anyone provide me some guidance?

No. Currently, you have $\displaystyle v,w \in \mathbb{R}^1$, so $\displaystyle v\cdot v = (a_1+a_2+\cdots+a_N)^2 \neq a_1^2+a_2^2+\cdots + a_N^2$

Instead, try $\displaystyle v = \left(a_1,a_2,\ldots,a_N\right), w = \left(\underbrace{\dfrac{1}{N},\dfrac{1}{N},\ldots ,\dfrac{1}{N}}_{N\text{ times}}\right)$. Now, $\displaystyle v\cdot w$ and $\displaystyle v\cdot v$ are as you wanted, but $\displaystyle w\cdot w = \dfrac{1}{N}$, not $\displaystyle \dfrac{1}{N^2}$. Fortunately, the inequality you are given has $\displaystyle \dfrac{1}{N}$ on the RHS, not $\displaystyle \dfrac{1}{N^2}$.
• Oct 23rd 2013, 08:33 AM
MichaelH
Re: Derive a relation using Cauchy Schwarz inequality
I do apologise, it was a typo on my end and I forgot to add that I was using 1/N n amount of times.

But was my calculation of $\displaystyle v$ dot product $\displaystyle v$ incorrect then?
• Oct 23rd 2013, 02:48 PM
SlipEternal
Re: Derive a relation using Cauchy Schwarz inequality
If you choose $\displaystyle v = a_1+a_2+\cdots + a_N$ and $\displaystyle w = \dfrac{1}{N}$, then those are real numbers. So, yes, your calculation of $\displaystyle v\cdot v$ is incorrect, but your calculation of $\displaystyle w\cdot w$ was correct. You would get $\displaystyle \dfrac{1}{N^2}$. You cannot take it $\displaystyle N$ times because you chose $\displaystyle w$ to be a real number, not a number repeated $\displaystyle N$ times.
• Oct 23rd 2013, 03:46 PM
MichaelH
Re: Derive a relation using Cauchy Schwarz inequality
{$\displaystyle v : \left(a_1,a_2,\ldots,a_N\right) | v \in R^N$} and {$\displaystyle w : \left(\underbrace{\dfrac{1}{N},\dfrac{1}{N},\ldots ,\dfrac{1}{N}}_{N\text{ times}}\right) | w \in R^N$}

Ok so I am still a little confused with whether I have this right or not now? And if this is correct for $\displaystyle v$ and $\displaystyle w$ then where do I go with $\displaystyle v \cdot v$, $\displaystyle w \cdot w$ and $\displaystyle v \cdot w$ in order to completely derive the inequality?
• Oct 23rd 2013, 03:56 PM
SlipEternal
Re: Derive a relation using Cauchy Schwarz inequality
Yes, you have it right now.

$\displaystyle v\cdot w = \dfrac{a_1}{N} + \dfrac{a_2}{N} + \cdots + \dfrac{a_N}{N} = \dfrac{a_1+a_2+\cdots + a_N}{N}$

$\displaystyle v\cdot v = a_1^2+a_2^2 + \cdots + a_N^2$

$\displaystyle w\cdot w = \underbrace{\dfrac{1}{N^2} + \dfrac{1}{N^2} + \cdots + \dfrac{1}{N^2}}_{N\text{ times}} = \dfrac{N}{N^2} = \dfrac{1}{N}$
• Oct 23rd 2013, 04:37 PM
MichaelH
Re: Derive a relation using Cauchy Schwarz inequality
Great. So now how do I show that the RHS is greater than or equal to the LHS using these facts?
• Oct 23rd 2013, 04:38 PM
SlipEternal
Re: Derive a relation using Cauchy Schwarz inequality
To quote you:

Quote:

Originally Posted by MichaelH
So I want to use $\displaystyle |v \cdot w|^2 =(v \cdot v)*(w \cdot w)$ to derive this.

Edit: Although, that should be $\displaystyle \le$, not $\displaystyle =$.
• Oct 23rd 2013, 04:41 PM
MichaelH
Re: Derive a relation using Cauchy Schwarz inequality
Quote:

Originally Posted by SlipEternal
To quote you:

Edit: Although, that should be $\displaystyle \le$, not $\displaystyle =$.

Ahh that's why I couldn't figure it out! Haha, my fault, should have double checked it. I was staring at = thinking... what am I supposed to do! Cheers mate.