About the equality in Cauchy-Schwarz

(i=1nai2)(i=1nbi2)(i=1naibi)2.\left(\displaystyle \sum_{i=1}^n a_i^2\right)\left( \displaystyle \sum_{i=1}^n b_i^2\right)\ge \left( \displaystyle \sum_{i=1}^n a_ib_i\right)^2.

Equality condition in Cauchy-Schwarz is aibi=k\frac{a_i}{b_i}=k for all i but what happens when bi=0{b_i} = 0. When the denominator is zero it should be undefined, but zero may be an equality solution.

E.g. a2+b2+c2ab+bc+ca a^2 + b^2 + c^2 \geq ab + bc + ca

By multiplying by 2 on both sides and a little rearrangement we get

(ab)2+(bc)2+(ca)20 (a-b)^2 +(b-c)^2 + (c-a)^2 \geq 0 Which is true for all reals a,b,c and also equality holds when a=b=c a=b=c even when a=b=c=0 a=b=c=0

We can also solve this by Cauchy-Schwarz :-

(a2+b2+c2)(b2+c2+a2)(ab+bc+ca)2(a^2 + b^2 + c^2)(b^2 + c^2 + a^2) \geq (ab + bc + ca)^2

Or, a2+b2+c2ab+bc+caa^2 + b^2 + c^2 \geq ab + bc + ca

And equality holds when,

ab=bc=ca=a+b+cb+c+a=1\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a} = 1

Therfore, we get, a=b=c a=b=c but a=b=c=0 a=b=c=0 will make the expression undefined but is also a valid solution.

Please help!!!

Note by Santu Paul
1 year ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Please explain your question from the start. Assume that the person reading this knows nothing about what you are writing.

Log in to reply

I have edited it and I think now it'll be easier to understand.

SANTU PAUL - 1 year ago

Log in to reply

Small correction at the end: setting a=b=c=0a=b=c=0 will not make the expression undefined, but rather indeterminate (there is a HUGE difference between the two terms mathematically). May I also suggest replacing the a,b,ca,b,c's with ai,bia_i,b_i's? It helps with the reading of the problem.

To answer your question directly (now that I have investigated the question after you have written up your question more clearly), if the bib_i's are zero, then equality still holds regardless; what you simply have is a degenerate case where the statement of when equality holds does not and cannot apply.

Log in to reply

@A Former Brilliant Member Can you please explain a little more clearly.

SANTU PAUL - 1 year ago

Log in to reply

@Santu Paul What did you want to clarify? If it is my answer to your question, then what I am saying that if the bib_i's are zero, then it's obvious that both sides of the equation will have to be zero, regardless of what the aia_i's are... so you still have equality of the Cauchy-Schwarz inequality. My point was that the condition for equality is contingent on the fact that none of the bib_i's are zero.

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...