Cauchy -schwarz inequality help

  • Context: Graduate 
  • Thread starter Thread starter dr hannibal
  • Start date Start date
  • Tags Tags
    Cauchy Inequality
Click For Summary

Discussion Overview

The discussion revolves around proving the Cauchy-Schwarz inequality, focusing on different methods of proof and understanding the notation involved in the expressions related to the inequality. Participants explore both algebraic expansions and the implications of specific summation indices.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant describes a method of proving the Cauchy-Schwarz inequality using the non-negativity of a quadratic expression.
  • Another participant seeks clarification on the notation used in the summation involving indices i and j, specifically regarding the terms that do not contribute when i equals j.
  • A participant explains that the summation is over indices where i and j differ, noting that terms where i equals j yield zero.
  • Further clarification is provided on the origin of the term -2a_ib_ja_jb_i in the expansion of the squared sum, suggesting that it arises from the subtraction of terms when i ≠ j.

Areas of Agreement / Disagreement

Participants generally agree on the mechanics of the summation and the expansion of the squared terms, but there is no consensus on the overall approach to proving the inequality or the clarity of the notation.

Contextual Notes

Participants express uncertainty regarding the notation and specific terms in the proof, indicating a need for further resources or examples to clarify these points.

Who May Find This Useful

Students and individuals studying mathematical inequalities, particularly those interested in the Cauchy-Schwarz inequality and its proofs.

dr hannibal
Messages
10
Reaction score
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
 
Last edited:
Physics news on Phys.org
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.
 
snipez90 said:
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 [tex]-2a_ib_ja_jb_i[/tex] coming from?
 
[tex](a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2[/tex]​

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.
 
Tedjn said:
[tex](a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2[/tex]​

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 threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K