# How to apply the Cauchy-Schwartz inequality on the following inner product?

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?

• \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? – user3482749 Nov 24 '18 at 18:02
• @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 – Hello Lili Nov 24 '18 at 18:12
• It does use the Cauchy-Schwartz inequality: that's precisely what that inequality is. – user3482749 Nov 24 '18 at 18:18
• @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. – Hello Lili Nov 24 '18 at 18:22
• @user3482749 I can give you a link if this helps: google.com/… – Hello Lili Nov 24 '18 at 18:24

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