Note this is for an inner product over $\mathbb{C}$.
I have the converse, but I am stuck at the first implication and in fact not even sure if it is true. Unless I am mistaken, I can reduce it to
$$(x\mid y) + (y\mid x) = 2\|x\|\|y\|$$
which implies that
$$\text{Re}((x\mid y)) = \|x\|\|y\|$$
since $(y\mid x)$ is the complex conjugate of $(x\mid y)$
but I am not sure how this helps me say that $x = ay$ or $y = ax.$
If it were $(x\mid y) = \|x\|\|y\|$ it would be true by Cauchy-Schwarz.
Thanks.
$$ \lVert x+y \rVert^2 = \lVert x \rVert^2 + \lVert y \rVert^2 + 2\Re\langle x , y \rangle \leq \lVert x \rVert^2 + \lVert y \rVert^2 + 2\lVert x \rVert\lVert y \rVert = (\lVert x \rVert+\lVert y \rVert)^2, $$ where the middle inequality is Cauchy–Schwarz, $$ \Re\langle x , y \rangle \leq \lvert \langle x, y \rangle \rvert \leq \lVert x \rVert \lVert y \rVert; $$ equality in the second occurs if and only if $y=ax$ or $x=ay$. To refine this to positive $a$, you can just check the case of equality for a more general $a$: the left-hand side is $\lvert 1+a \rvert $ times the norm of $x$ (or $y$), while the right is $1+\lvert a \rvert$ times the appropriate norm, and equality holds if and only if $a>0$. Or, you can look at the inequality $\Re\langle x , y \rangle \leq \lvert \langle x, y \rangle \rvert$ and work out when that attains equality.