- #1
dr hannibal
- 10
- 0
Need help proving Cauchy Schwarz inequality ...
the first method I know is pretty easy
[itex]\displaystyle\sum_{i=1}^n (a_ix-b_i)^2 \geq 0 [/itex]
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..
[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 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 :) ...[itex]\displaystyle\sum_{i\not=j}^n ((a_i)^2(b_j)^2+(a_j)^2(b_i)^2-2a_ib_ja_jb_i ) [/itex]Thanks
the first method I know is pretty easy
[itex]\displaystyle\sum_{i=1}^n (a_ix-b_i)^2 \geq 0 [/itex]
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..
[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 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 :) ...[itex]\displaystyle\sum_{i\not=j}^n ((a_i)^2(b_j)^2+(a_j)^2(b_i)^2-2a_ib_ja_jb_i ) [/itex]Thanks
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