Cauchy -schwarz inequality help

1. Jul 5, 2010

dr hannibal

Need help proving Cauchy Schwarz inequality ...

the first method I know is pretty easy

$\displaystyle\sum_{i=1}^n (a_ix-b_i)^2 \geq 0$

expanding this and using the discriminatant quickly establishes the inequality..

The 2nd method I know is I think a easier one , but I dont have a clue about how this notation works..

Since cauchy SHwarz inquality states..

$$(a_1b_1+a_2b_2+...+a_nb_n)^2 \leq ((a_1)^2+(a_2)^2+..+(a_n)^2)((b_1)^2+(b_2)^2+...+(b_n)^2)$$

$$((a_1)^2+(a_2)^2+..+(a_n)^2)((b_1)^2+(b_2)^2+...+(b_n)^2)-(a_1b_1+a_2b_2+...+a_nb_n)^2 \geq 0$$

I dont usnderstand how the below notation works as I cant follow from the above line to the line below , if someone can point me to some resources where I can know more about it :) ...

$\displaystyle\sum_{i\not=j}^n ((a_i)^2(b_j)^2+(a_j)^2(b_i)^2-2a_ib_ja_jb_i )$

Thanks

Last edited: Jul 5, 2010
2. Jul 5, 2010

snipez90

That's just summing over the quantities within the parentheses for which the two indices i and j differ. Note that if i = j, the quantity inside the summation is just 0, so it does not contribute to the sum.

3. Jul 5, 2010

dr hannibal

thanks :) , one more small question when they have summed the above where is $$-2a_ib_ja_jb_i$$ comming from?

4. Jul 5, 2010

Tedjn

$$(a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$$​

Expanded out, convince yourself that whenever i ≠ j you get a 2aibiajbj. To do this, it helps to write out a simple case for small n (n = 2 or 3) and look at how what terms you get that involve i ≠ j. The sign in your example is negative, because you are subtracting.

5. Jul 5, 2010

thanks