Inner product inequality sort of?

**davismj** · Nov 5th 2009, 01:45 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** · Nov 5th 2009, 02:09 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** · Nov 5th 2009, 08:42 PM

Originally Posted by Jameson

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

$\sum_{j=1}^n j(a_j){b_j \over j}<br />$

**Jameson** · Nov 5th 2009, 09:28 PM

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

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