Cauchy -schwarz inequality help

dr hannibal

Need help proving Cauchy Schwarz inequality ...

the first method I know is pretty easy

[latex]\displaystyle\sum_{i=1}^n (a_ix-b_i)^2 \geq 0 [/latex]

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..

[tex](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)[/tex]

[tex]((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[/tex]

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 :) ...


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


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

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

Tedjn

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

thanks