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
4 weeks ago

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

- 4 weeks ago

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

- 3 weeks, 6 days 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.

- 3 weeks, 6 days ago

Can you please explain a little more clearly.

- 3 weeks, 6 days 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.

- 3 weeks, 6 days ago