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

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richyw
 2 years ago
Best ResponseYou've already chosen the best response.0The 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.

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0first 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?

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Usually really weird stuff like that comes about by doing things backwards.

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

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Like what if \(\mathbf{b} = 0\)

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

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1If 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.

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0does that show that this is the only case though?

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0like the "and only" part of if and only if

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1Thats 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.
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