Prove $ | \sum_{k=1}^{n} \frac{z_k}{k} |^2 \le ( \sum_{k=1}^{n} \frac{1}{k^2} ) (\sum_{k=1}^{n}| z_k|^2 ) $

Question

I want to prove that: $ \left| \sum_{k=1}^{n} \frac{z_k}{k} \right|^2 \le \left( \sum_{k=1}^{n} \frac{1}{k^2} \right) \cdot \left(\sum_{k=1}^{n} \left| z_k\right|^2 \right) $
where $z_k \in \mathbb C $
Probably I should use Schwarz inequality:
$$\left| \left\langle x,y \right\rangle \right| \le \left| \left| x \right| \right| \cdot \left| \left| y \right| \right| = \sqrt{\left\langle x,x \right\rangle \cdot \left\langle y,y \right\rangle} $$ after square $$ \left| \left\langle x,y \right\rangle \right|^2 \le \left\langle x,x \right\rangle \cdot \left\langle y,y \right\rangle $$
I thought that I can: $$ \vec{x} = [\frac{1}{1^2},\frac{1}{2^2},...,\frac{1}{n^2}]^T $$ $$ \vec{y} = [\left| z_1\right|^2,...,\left| z_n\right|^2]^T $$ and put $$ \left\langle x,y\right\rangle = \sum_{k=1}^{n} x_k \cdot y_k $$ Right side I get instantly. The problem is with left side: $$ \left| \sum_{k=1}^{n} \frac{|z_k|^2}{k^2} \right|^2 = \left| \sum_{k=1}^{n} \left| \frac{z_k^2}{k^2} \right| \right|^2 = \left| \sum_{k=1}^{n} \frac{z_k^2}{k^2} \right|^2 $$ but it is not$$ \left| \sum_{k=1}^{n} \frac{z_k}{k} \right|^2 $$ And I have 2 questions:
1. Can I define that: $ \left\langle x,y\right\rangle = \sum_{k=1}^{n} x_k \cdot y_k $
2. Where is the mistake? / How can I finish that?

Answer 1

With your choice of $x$ and $y$ the right-hand side would be $$ \left( \sum_{k=1}^{n} \frac{1}{k^4} \right) \cdot \left(\sum_{k=1}^{n} \left| z_k\right|^4 \right) \, . $$

But applying Cauchy-Schwarz to $\vec x = (z_1, \ldots, z_n)^T$ and $\vec y = (1, \frac 12,\ldots \frac 1n)^T$ with the standard complex inner product $$ \left\langle \vec x, \vec y\right\rangle = \sum_{k=1}^{n} x_k \cdot \bar y_k $$ gives the intended result.