Proof: $x, y$ linear depending $ \Leftrightarrow|\langle x,y\rangle| = ||x|| ||y|| $ Thus far:$x, y$ linear depending $\Rightarrow \exists r \in \mathbb{R} : x = r\cdot y $
Therefore on gets: $ |\langle x,y\rangle| = r \cdot \sum\limits_{i = 1}^{n} x_{i}^2 = \sqrt{ ( \sum\limits_{i = 1}^{n} x_{i}^2 ) \cdot \sum\limits_{i = 1}^{n} (rx_{i})^2 } = \sqrt{ (r^2 \cdot ( \sum\limits_{i = 1}^{n} x_{i}^2 ) \cdot ( \sum\limits_{i = 1}^{n} x_{i}^2) } = r \cdot \sum\limits_{i = 1}^{n} x_{i}^2$
The other direction however makes me trouble: 
I know: $\cos(\alpha) = \frac{|\langle x,y\rangle|}{||x|| ||y|| } = 1$ if and only if $\alpha = 0$ for $ 0 \le \alpha <2\pi $
However using this argument feels like mere shifting of the problem, because how does one know that $\alpha $  describes the angle between two vectors and that they are linear depending? This sounds like an abuse of definition + I introduce something like trigonometry, even though it should not be needed for this  problem. 
As in most cases the wood cutter's method is guaranteed to produce a result and thus tried to go the long way: 
$ \sqrt{ (\sum\limits_{i = 1}^{n} x_{i}^2) \cdot ( \sum\limits_{i = 1}^{n} y_{i}^2 )} $
$ = \sqrt{\sum\limits_{i = 1}^{n} (x_{i}^2 \cdot \sum\limits_{j = 1}^{n} y_{j}^2 )  }  $
But now I am stuck.I also thought about factor everything out and get a huge sum and look, if something could become neater, but that also didn't seem very promising. 
Did I overlook something? I would very happy, if someone could point out, where I had gone wrong or just what I should have done better. 
As always any constructive comment, answer is appreciated.  

The fact that one vector is a multiple of the other [b]tells[/b] you that the angle between them is 0 and so has cosine 1.  Equivalently, the cosine of the angle between the two vectors is 1 means that the angle between them is 0 and one is a multiple of the other.
