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Under what conditions on \(\mathbf{a}\) and \(\mathbf{b}\) is Cauchy's inequality an equality? Examine the proof

Mathematics
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The proof is done as shown. We introduce the real variable\(t\) and consider the function \[f(t)=|\mathbf{a}-t\mathbf{b}|^2=|\mathbf{a}|^2-2t\mathbf{a}\cdot\mathbf{b}+t^2|\mathbf{b}|^2\] This is a quadratic function of t. Its minimum value occurs at \[t=\frac{(\mathbf{a}\cdot\mathbf{b})}{|\mathbf{b}|^2}\] and that minimum value is \[f\left((\mathbf{a}\cdot\mathbf{b})/|\mathbf{b}|^2\right)=|\mathbf{a}|^2-\frac{(\mathbf{a}\cdot\mathbf{b})}{|\mathbf{b}|^2}\]clearly \(f(t)\geq0\,\forall \,t\) so \[|\mathbf{a}|^2-\frac{(\mathbf{a}\cdot\mathbf{b})}{|\mathbf{b}|^2}\geq 0\] which completes the proof.
first of all. I have no idea why they did the proof like that. How would I ever guess to do that? second, how to I examine this proof to find where cauchy's inequality an equality?
Usually really weird stuff like that comes about by doing things backwards.

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Other answers:

To answer your second question... look for things like division by variables to see where holes would be.
Like what if \(|\mathbf{b}| = 0\)
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does anyone know how I would show that it is an equality if an only if a and b are colinear?
If you assume that a and b are colinear, then:\[a=tb\]for some real number t. If you plug this in the cauchy inequality it turns out to be an equality.
does that show that this is the only case though?
like the "and only" part of if and only if
Thats correct, what I posted above is only one direction of the proof (if a and b are colinear, then there is equality in Cauchy's inequality). For the other direction, you should do a proof by contradiction. Assume that there is equality, and that a and b are not colinear. Since a and b are not colinear, you have that the function:\[f(t)=|a-tb|^2\] is always strictly greater than 0 for any value of t. So using:\[|a-tb|^2>0\]and \[|(a\cdot b)|=|a|\cdot |b|\] you should end up with a contradiction.

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