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 ) $ 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?

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.
