Suppose $(H,(\cdot,\cdot))$ is a Hilbert space, $||\cdot||$ is the norm induced by the scalar product. I am trying to find all vectors that satisfy $$||x+y|| = ||x||+||y||$$ I think equality holds if and only if $x = \lambda y$.
My attempt:
If $x = \lambda y$, then by plugging in directly, we get equality of the left and right side (assuming $\lambda$ is real).
Conversely, if $||x+y||^2 = (||x||+||y||)^2$ then $(x+y,x+y) = (x,x)+(y,y)+2(x,x)(y,y)$ and so $(x,y)+(y,x) = 2(x,x)(y,y)$. Now I want to continue by saying that $2(x,y) = 2(x,x)(y,y)$ so by Cauchy-Schwarz equality occurs when $x = \lambda y$. However, this would not work in a complex Hilbert space, since I cannot then say that $(x,y)+(y,x)$ is the same as $2(x,y)$. How should I proceed?
In the complex case, you can still say that $(x,y)+(y,x)=(x,y)+\overline{(x,y)}=2\operatorname{Re}(x,y)$. Since $|\operatorname{Re}(x,y)|\leq |(x,y)|$, you can still use Cauchy-Schwarz to get that $|(x,y)+(y,x)|\leq 2\|x\|\|y\|$, and that in order for equality to hold $x$ must be a scalar multiple of $y$ (assuming $y\neq 0$).
(Note that in both the real and complex case, however, Cauchy-Schwarz in the way this argument uses it only tells you $x=\lambda y$ for some scalar $\lambda$ is necessary, not that it is sufficient. In fact, it is not sufficient in either case. A simple way to figure out what restrictions there are on $\lambda$ is just to plug in $y=\lambda x$ and see what $\|x+y\|=\|x\|+\|y\|$ tells you. Note that contrary to what you claim, it is not sufficient for $\lambda$ to be real: think about what happens when $\lambda=-1$, for instance.)