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

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

E.g.\( a^2 + b^2 + c^2 \geq ab + bc + ca \)

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

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

We can also solve this by Cauchy-Schwarz :-

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

Or, \(a^2 + b^2 + c^2 \geq ab + bc + ca\)

And equality holds when,

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a} = 1\)

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

Please help!!!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestPlease 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.

Log in to reply

Small correction at the end: setting \(a=b=c=0\) 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 \(a,b,c\)'s with \(a_i,b_i\)'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 \(b_i\)'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

Log in to reply

Log in to reply