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

Question

richyw · Answer

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.

richyw · Answer

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?

anonymous · Answer

Usually really weird stuff like that comes about by doing things backwards.

anonymous · Answer

To answer your second question... look for things like division by variables to see where holes would be.

anonymous · Answer

Like what if $|\mathbf{b}| = 0$

experimentx · Answer

attachment

richyw · Answer

does anyone know how I would show that it is an equality if an only if a and b are colinear?

anonymous · Answer

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.

richyw · Answer

does that show that this is the only case though?

richyw · Answer

like the "and only" part of if and only if

anonymous · Answer

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.