## Cauchy -schwarz inequality help

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
 PhysOrg.com mathematics news on PhysOrg.com >> Pendulum swings back on 350-year-old mathematical mystery>> Bayesian statistics theorem holds its own - but use with caution>> Math technique de-clutters cancer-cell data, revealing tumor evolution, treatment leads
 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.

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

## Cauchy -schwarz inequality help

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

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.

 Quote by 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.
thanks

 Similar discussions for: Cauchy -schwarz inequality help Thread Forum Replies Calculus & Beyond Homework 1 General Math 10 Linear & Abstract Algebra 6 Calculus & Beyond Homework 1 Calculus & Beyond Homework 4