Cauchy -schwarz inequality help

[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 don't 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 don't usnderstand how the below notation works as I can't 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

 

[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.

 

[dr hannibal]

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.

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

 

[Tedjn]

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

Expanded out, convince yourself that whenever i ≠ j you get a 2a_ib_ia_jb_j. 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.

 

[dr hannibal]

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

Expanded out, convince yourself that whenever i ≠ j you get a 2a_ib_ia_jb_j. 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.

thanks