# About the equality in Cauchy-Schwarz

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

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.

Note by Santu Paul
1 year, 3 months ago

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- 1 year, 3 months ago

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

- 1 year, 3 months ago

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.

- 1 year, 3 months ago

Can you please explain a little more clearly.

- 1 year, 3 months ago

What did you want to clarify? If it is my answer to your question, then what I am saying that if the $b_i$'s are zero, then it's obvious that both sides of the equation will have to be zero, regardless of what the $a_i$'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 $b_i$'s are zero.

- 1 year, 3 months ago