TITLE: Triangle equality implies vector dependence.
QUESTION [5 upvotes]: I am trying to prove this statement:

Show that if $x$ and $y$ are two vectors in an inner product space such that $||x+y||=||x||+||y||$, then $x$ and $y$ are linearly dependent. 

Squaring the equality I get
$$\langle x+y,x+y\rangle=\langle x,x\rangle +2||x||\cdot||y||+\langle y,y\rangle $$
then, using linearity of the inner product I get
$$ \langle x,x\rangle +\langle y,y\rangle+\langle x,y\rangle+\langle y,x\rangle=\langle x,x\rangle +2||x||\cdot||y||+\langle y,y\rangle $$
After all the cancellation I finally arrive at
$$ \mathrm{Re}\langle x,y\rangle=||x||\cdot||y|| $$
This looks like Cauchy-Schwarz inequality, so the only thing left to show is that $\mathrm{Re}\langle x,y\rangle=|\langle x,y\rangle|$, how can I do that?

REPLY [3 votes]: Note that through an application of Cauchy-Schwarz we get
$$\rm{Re}\langle \mathbf{x},\ \mathbf{y}\rangle = \|\mathbf{x}\|\|\mathbf{y}\|\ge|\langle \mathbf{x},\ \mathbf{y}\rangle|$$
This is only possible if there is equality since we naturally have 
$$\rm{Re}\langle \mathbf{x},\ \mathbf{y}\rangle \le|\langle \mathbf{x},\ \mathbf{y}\rangle|$$