# Proving a Summation Inequality

I've gotten stuck with this homework problem ...

Prove that $$(\sum_{j=1}^{n} a_jb_j)^2 \le (\sum_{j=1}^{n} j a_j^2)(\sum_{j=1}^{n} \frac{b_j^2}{j})$$ for all real numbers $a_1, ... , a_n$ and $b_1, ... , b_n$.

Here's what I have so far. Let's make two vectors $a = (a_1, ... , a_n), b = (b_1, ... , b_n) \in \mathbb{R}^n$. Then we can express the left side in terms of an inner product: $$\sum_{j=1}^{n} a_jb_j = \langle a, b \rangle$$ I suspect that this is the correct first step, since we are learning about inner product spaces. Now we can also rewrite the right side, but less neatly ... $$\sum_{j=1}^{n} j a_j^2 = a_1^2 + 2a_2^2 + \ldots + na_n^2 = \langle a, a \rangle + \sum_{j=1}^n (j - 1)a_j^2$$ $$\sum_{j=1}^{n} \frac{b_j^2}{j} = b_1^2 + \frac{b_2^2}{2} + \ldots + \frac{b_n^2}{n} = \langle b, b \rangle + \sum_{j = 1}^n -\frac{(j-1)b_j^2}{j}$$

So our original equation reduces to $\langle a, b \rangle^2 \le \langle a, a \rangle \langle b, b\rangle + stuff$. By the Cauchy-Schwarz inequality we know that $\langle a, b \rangle^2 \le \langle a, a \rangle \langle b, b\rangle$ but the right side still has other stuff going on. It would be nice to prove that the $stuff$ is non-negative, but I can't see how to do this (or even if it's necessarily true).

Thanks for any help!

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If you know that $$\left(\sum_{j=1}^n c_j d_j\right)^2 \le \left(\sum_{j=1}^n c_j^2\right)\left(\sum_{j=1}^n d_j^2\right)$$ then consider the case where $c_j= ja_j$ and $d_j = \dfrac{b_j}{j}$.