# Proving inner-product inequality

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.

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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

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 $$\left(\sum_{k=1}^nc_kd_k\right)^2\le\left(\sum_{k=1}^nc_k^2\right)\left(\sum_{k=1}^nd_k^2\right)\;.$$