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

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richyw Group TitleBest ResponseYou've already chosen the best response.0
The proof is done as shown. We introduce the real variable\(t\) and consider the function \[f(t)=\mathbf{a}t\mathbf{b}^2=\mathbf{a}^22t\mathbf{a}\cdot\mathbf{b}+t^2\mathbf{b}^2\] 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.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Usually really weird stuff like that comes about by doing things backwards.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
To answer your second question... look for things like division by variables to see where holes would be.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Like what if \(\mathbf{b} = 0\)
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
does anyone know how I would show that it is an equality if an only if a and b are colinear?
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
does that show that this is the only case though?
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
like the "and only" part of if and only if
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
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:\[f(t)=atb^2\] is always strictly greater than 0 for any value of t. So using:\[atb^2>0\]and \[(a\cdot b)=a\cdot b\] you should end up with a contradiction.
 one year ago
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