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
2 years, 6 months ago

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1 vote

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

A Former Brilliant Member - 2 years, 6 months ago

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

SANTU PAUL - 2 years, 6 months ago

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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.

A Former Brilliant Member - 2 years, 5 months ago

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

SANTU PAUL - 2 years, 5 months ago

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@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.

A Former Brilliant Member - 2 years, 5 months ago

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