Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

To answer your second question... look for things like division by variables to see where holes would be.
Like what if \(|\mathbf{b}| = 0\)
1 Attachment
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question