Given an inner product space X, let x,y $\in$ X. I need to show the following are equivalent:
$\|$x+y$\|$ = $\|$x$\|$ + $\|$y$\|$ $\iff$ $\|$y$\|$x=$\|$x$\|$y .
I know the first condition makes x and y linearly dependent but I haven't gone much further than the Cauchy-Schwarz inequality, triangle inequality and parallelogram law so I'm not sure how to get the "free" x and y on the right hand side.
If $x$ is the zero vector, the equivalence holds. Suppose that $x$ is nonzero. Rewrite the left-hand side of the equivalence as $$ \sqrt{(x+y,x+y)}=\sqrt{(x,x)}+\sqrt{(y,y)} $$ and raise both parts to the square, thereby getting that $$ (x,y)=|x|\,|y|, $$ or \begin{equation*} \tag{1} (x/|x|,y)=|y|. \end{equation*} Consider then an orthonormal basis $x_1,\ldots,x_n$ such that $x_1=x/|x|$ and let $$ y=\alpha_1 x_1 + \ldots + \alpha_n x_n. $$ Then it follows from (1) that $$ \alpha_1 = \sqrt{\alpha_1^2 + \ldots +\alpha_n^2}, $$ whence $$ \alpha_1=|y|, \alpha_2=\ldots=\alpha_n=0. $$ or $$ y=|y| x/|x| \iff |x| y = |y| x. $$ The converse is trivial.
The stated condition is equivalent to the following conditions $$ \|x+y\|^2 = (\|x\|+\|y\|)^2 \\ \|x\|^2+2\Re(x,y)+\|y\|^2=\|x\|^2+2\|x\|\|y\|+\|y\|^2 \\ \Re(x,y) = \|x\|\|y\|. $$ And the following are also equivalent $$ \|\|y\|x-\|x\|y\|^2 = 0 \\ \|y\|^2\|x\|^2-2\|x\|\|y\|\Re(x,y)+\|x\|^2\|y\|^2 = 0 \\ 2\|x\|\|y\|\{\|x\|\|y\|-\Re(x,y)\}=0. $$ If $x=0$ or $y=0$ then $\|x\|y=\|y\|x$ and $\|x+y\|=\|x\|+\|y\|$. Otherwise, if neither is $0$, then $$ \|x+y\| = \|x\|+\|y\| \iff \|y\|x-\|x\|y = 0. $$
There is no need to use bases. ||x+y|| = ||x||+||y|| implies x²+y²+2x•y = x²+y²+2||x||||y||, so 2x•y=2||x||||y||. Now write x=||x||*y/||y||+v, where v is the "error term". This becomes
2||x||y•y/||y|| + 2v•y = 2||x||||y||
So 2v•y = 0. Reversing that y=(x-v)*||y||/||x|| shows that v•x=0 as well. This means that v is in space of vectors normal to x and y. At the same time, v is in the subspace spanned by x and y. These are complementary, so v=0; done.