A community for students.
Here's the question you clicked on:
 0 viewing
richyw
 3 years ago
Under what conditions on \(\mathbf{a}\) and \(\mathbf{b}\) is Cauchy's inequality an equality? Examine the proof
richyw
 3 years ago
Under what conditions on \(\mathbf{a}\) and \(\mathbf{b}\) is Cauchy's inequality an equality? Examine the proof

This Question is Closed

richyw
 3 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
 3 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Usually really weird stuff like that comes about by doing things backwards.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To answer your second question... look for things like division by variables to see where holes would be.

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

richyw
 3 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?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thats 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.