
#1
Jul510, 08:34 AM

P: 10

Need help proving Cauchy Schwarz inequality ...
the first method I know is pretty easy [latex]\displaystyle\sum_{i=1}^n (a_ixb_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)^22a_ib_ja_jb_i ) [/latex] Thanks 



#2
Jul510, 09:05 AM

P: 1,106

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
Jul510, 09:31 AM

P: 10





#4
Jul510, 12:20 PM

P: 740

Cauchy schwarz inequality help[tex](a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2[/tex] Expanded out, convince yourself that whenever i ≠ j you get a 2a_{i}b_{i}a_{j}b_{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. 



#5
Jul510, 12:45 PM

P: 10




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