Cauchy-Schwarz inequality for points on unit sphere?

Question

I have the following problem (I added a photo of the problem):

For a set of points $x \in \pi_k$ with $x \in R^d$ and on unit sphere we compute:

$m_j = \frac{1}{n_j}\sum_{x \in \pi_k} x $

and normalize

$c_j = \frac{m_j}{||m_j||}$.

Then for any unit vector $z \in R^d$ we use the Cauchy-Scharz inequality

$ \sum_{x \in \pi_k$} x^Tz \leq \sum_{x \in c_j} x^Tc_j $

to see that $c_j$ can be seen as the point to which (in an average sense) all $x \in \pi_j$ are most cosine-similar.

Can somebody explain why this hold? I am confused because I don't see where Cauchy-schwarz is used because $z$ and $c_j$ are different?

Answer 1

Since $m_j$ is the average of $x$ over $\pi_k$ and $c_j$ is the normalised form of $m_j$, the sum $\sum_{x\in\pi_k}x^T$ is actually a positive scalar multiple of $c_j$.

Let $\sum_{x\in\pi_k}x^T=pc_j$, where $p>0$. The inequality $\sum_{x\in\pi_k}x^Tz\le\sum_{x\in\pi_k}x^Tc_j$ can then be rewritten as $pc_j^Tz\le pc_j^Tc_j$. That is, $c_j^Tz\le1$. This is clearly true because both $c_j$ and $z$ are unit vectors.