Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Homework problem.

Let $a_1, a_2, ..., a_n$ and $b_1,b_2,...,b_n$ be sets of real numbers. Show that: $$ \left(\sum_{k=1}^n a_kb_k\right)^2 \leq \left(\sum_{k=1}^n ka_k^2\right) \left(\sum_{k=1}^n\frac{b_k^2}{k}\right)$$

for all $n \geq 1$.


The hint given to us was not to prove this with induction, but to think of the problem "in linear algebra terms".

I've pondered this for a few days now, and come up with this: You can think of the $a$'s as a vector $\langle a_1,...,a_n\rangle$, and the $b$'s as a vector $\langle b_1,...,b_n\rangle$ and then the problem can be rephrased as inner products: $$\langle A,B\rangle\langle A,B\rangle \space \leq \space \langle A,A\rangle\langle K^{-1}B,B\rangle\;,$$

where $A$ and $B$ are defined above and $KA$ is $\langle 1a_1, 2a_2, ..., na_n\rangle$ and $K^{-1}B$ is $\langle 1b_1, \frac{1}{2}b_2,...,\frac{1}{n}b_n\rangle$.

I'm aware of the similarity with the Cauchy-Schwarz inequality, but can't figure out how to manipulate what I have any further.

Any insights are appreciated.

share|improve this question
4  
Hint: $a_k b_k = \sqrt{k} a_k \frac{b_k}{\sqrt{k}}$. –  copper.hat May 4 '12 at 3:02
    
Thanks so much for the hint, that really opened it up for me! –  Daniel May 4 '12 at 3:17
add comment

1 Answer

up vote 1 down vote accepted

HINT: Let $c_k=a_k\sqrt k$ and $d_k=\frac{b_k}{\sqrt k}$. The rest is spoiler-protected; mouse-over to see it.

Then $$\left(\sum_{k=1}^n a_kb_k\right)^2 =\left(\sum_{k=1}^nc_kd_k\right)^2\;,$$ and $$\left(\sum_{k=1}^n ka_k^2\right) \left(\sum_{k=1}^n\frac{b_k^2}{k}\right)=\left(\sum_{k=1}^nc_k^2\right)\left(\sum_{k=1}^nd_k^2\right)\;,$$ and your problem is to show that $$\left(\sum_{k=1}^nc_kd_k\right)^2\le\left(\sum_{k=1}^nc_k^2\right)\left(\sum_{k=1}^nd_k^2\right)\;.$$

share|improve this answer
    
Thanks, the first line of yours and copper.hat's hint really opened it up for me! Check marking your answer, do I need to close this question somehow? –  Daniel May 4 '12 at 3:18
    
@Precision: Nope, you've done everything. Including the best part: solving the problem! :-) –  Brian M. Scott May 4 '12 at 3:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.