# Thread: Cauchy-Schwartz Inequality

1. ## Cauchy-Schwartz Inequality

This is a question that has been bugging me for quite some time. When I was a sophomore in Linear Algebra the Professor gave us a proof of the Cauchy Schwarz inequality like so,
For any vectors u,v in R^n
| u*v | = | ||u||*||v|| * cos (theta)|
= ||u|| * ||v|| * |cos(theta)|
<= ||u|| * ||v|| * 1 since |cos(theta)|<=1
= ||u|| * ||v||

Now here is my question, the first step of this proof requires knowing that u*v= ||u||*||v||*cos(theta), which is easily proven using the Law of Cosines. But, the Law of Cosines doesn't necessarily work in R^n for n>3. Therefore, this proof only works for the case when n<=3. Am I correct? Or does the first step of the proof work in any dimension? I'm trying to make the transition from R^n to general normed vector spaces and I gotta get this stuff clear in my head!! Thanks for any replies.

2. Here's a proof that doesn't require the law of cosines.

Consider the expression f(t) = |u + tv|^2 = (u+tv)*(u+tv). Obviously, f(t) >=0 for all t.

Expand:
f(t) = u*u + tu*v + tv*u + t^2v*v
= |u|^2 + 2tu*v + t^2|v|^2
= c + bt + at^2

with c = |u|^2, b = 2u*v, a = |u|^2.

This is a quadratic function that is never negative (has at most one real zero).

Thus the discriminant, b^2-4ac is non-positive, b^2-4ac <=0 and thus
4(u*v)^2 < 4|u|^2|v|^2.
Divide by 4 and take the square root, and you get the result:

|u*v| <= |u||v|.

The law of cosines (in any dimension) then is a consequence, or rather you can define the angle theta between two vectors u,v by u*v = |u||v|cos(theta).

Hope this helps

3. Thank you, that's the first time I ever saw that version of the proof.