Inner product inequality sort of?

davismj · #1 $Old$ November 5th, 2009, 03:45 PM

Okay. The left side is clearly <a,b>. The right side is...um?

Let j = (1,2,...,n) and k = (1,1/2,...,1/n), then <sqrt(j),a><sqrt(k),b>.

I guess? I don't have any idea what to do with this.

Jameson · #2 $Old$ November 5th, 2009, 04:09 PM

Since the indices are the same the scalar j can cancel out as every term of the first sum will have (1/j) and every term of the second will have a j. It is simply the Cauchy-Schwarz inequality now.

Cauchy?Schwarz inequality - Wikipedia, the free encyclopedia

davismj · #3 $Old$ November 5th, 2009, 10:42 PM

Quote:

Originally Posted by Jameson $View Post$

Since the indices are the same the scalar j can cancel out as every term of the first sum will have (1/j) and every term of the second will have a j. It is simply the Cauchy-Schwarz inequality now.

Cauchy?Schwarz inequality - Wikipedia, the free encyclopedia

I'm not sure that this holds. Notice the difference between:

$\sum_{j=1}^n j(a_j) \sum_{j=1}^n {b_j \over j}$

and

$\sum_{j=1}^n j(a_j){b_j \over j}$

Jameson · #4 $Old$ November 5th, 2009, 11:28 PM

Yes, of course. Got too excited to post a cool link. Won't work.