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Reading a research article, I came across the following statement:

The following function (where $x_i$ and $p_j$ are two vectors in $\mathbb{R}^n$ and $\mu_{i,j}$ is a constant) - it doesn't matter where it came from, is just an inner product actually:

$\sum_{i} \mu_{i,j} - \left< \sum_{i} \mu_{i,j} \frac{x_i}{||x_i||}, \frac{p_j}{||p_j||} \right>$

is minimized by using the Cauchy-Schwartz inequality iff

$p_j \propto \sum_{i} \mu_{i,j} \frac{x_i}{||x_i||}$.

I want to apply the Cauchy-Schwartz inequality on $\left< \sum_{i} \mu_{i,j} \frac{x_i}{||x_i||}, \frac{p_j}{||p_j||} \right>$ in order to understand where the proportionality thing came from.

So basically the Cauchy-Schwartz inequality states that: $\left< x, y\right> \le ||x|| \cdot ||y||$. But how can I apply this formula on that monstrous term?

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    $\begingroup$ \begin{align*}\left\langle \sum_i\mu_{i,j}\frac{x_i}{\|x_i\|},\frac{p_j}{\|p_j\|}\right\rangle &= \sum_i\mu_{i,j}\left\langle\frac{x_i}{\|x_i\|},\frac{p_j}{\|p_j\|}\right\rangle \\&= \sum_i\mu_{i,j}\frac{\langle x_i,p_j\rangle}{\|x_i\|\|p_j\|}\\&\leq\sum_i\mu_{i,j} \end{align*} Does that help? $\endgroup$ – user3482749 Nov 24 '18 at 18:02
  • $\begingroup$ @user3482749 no, this is actually where the formula I stated came from (like going backwards), and it doesn't use the Cauchy-Schwartz inequality, but thanks for your contribution anyway $\endgroup$ – Hello Lili Nov 24 '18 at 18:12
  • $\begingroup$ It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is. $\endgroup$ – user3482749 Nov 24 '18 at 18:18
  • $\begingroup$ @user3482749 I don't think this is the correct answer because as I said, in the article the equations you stated are used to generate the formula in my question. So it's like going backwards. $\endgroup$ – Hello Lili Nov 24 '18 at 18:22
  • $\begingroup$ @user3482749 I can give you a link if this helps: google.com/… $\endgroup$ – Hello Lili Nov 24 '18 at 18:24
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Beyond the inequality itself, the full Cauchy-Schwarz theorem states that $|\langle u, v\rangle|$ is maximized exactly when $u$ and $v$ are parallel or anti-parallel. I.e., when $v \propto u$. For your inner product, this is

$$\frac{p_j}{\|p_j\|} \propto \sum_{i} \mu_{i,j} \frac{x_i}{||x_i||}$$

Since $1/\|p_j\|$ is just a scalar multiplier, it can be absorbed into the constant of proportionality, leaving $$p_j \propto \sum_{i} \mu_{i,j} \frac{x_i}{||x_i||}$$

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