How to prove that $\lVert x + y\rVert = \lVert x\rVert + \lVert y \rVert \implies \lVert tx + (1-t)y\rVert = t\lVert x\rVert + (1-t)\lVert y \rVert$? If a norm on a vector space is given by an inner product then we have $\lVert x + y\rVert = \lVert x \rVert + \lVert y \rVert$ only when $x=\lambda y$ for some constant $\lambda$ (this follows from the Cauchy-Schwarz inequality).
This does not hold for general (unitary invariant) norms.
E.g. when considering generalised Ky Fan norms:
$$ \lVert x\rVert_s = \sum_i s_i \lvert x\rvert_i^\downarrow$$
where $s_i\geq 0$ and $\lvert x\rvert^\downarrow$ is $\lvert x\rvert$ rearranged with the components from biggest to smallest. So for example the operator (sup) norm has $s_1 = 1$ and $s_i=0$ for $i>1$, while the trace norm has $s_i=1$ for all $i$. For these norms it is in general possible to find $x$ and $y$ so that the norm is additive for these vectors, while $x$ and $y$ are not multiples of each other.
Now my question is the following. Given a unitarily invariant norm and vectors $x$ and $y$ such that $\lVert x + y\rVert = \lVert x\rVert + \lVert y \rVert$, can we then conclude that $\lVert tx + (1-t)y\rVert = t\lVert x\rVert + (1-t)\lVert y \rVert$ for $0\leq t \leq 1$? 
This is true for these generalised Ky Fan norms and for all norms that come from inner products. It is also true for all $p$-norms as this additivity still only holds for $x$ and $y$ that are multiples of each other (even though $p$-norms don't come from an inner product).
My intuition behind this is that this additivity property (when $x$ and $y$ aren't multiples of each other) forces the norm to be "linear" in a certain way, such that the norm has to be very similar to a generalised Ky Fan norm. So as a bonus question: If there are $x$ and $y$ that aren't multiples of each other, such that the norm is additive for these $x$ and $y$, show that the norm is a generalised Ky Fan norm. Note that this can only be true when we assume the norm to be unitary invariant since otherwise we can construct a counter-example of this (take $(X_1,\lVert\cdot\rVert_1)\oplus (X_2,\lVert\cdot\rvert_2)$ with the norm $\sup\{\lVert\cdot\rVert_1,\lVert\cdot\rVert_2\}$ where $\lVert\cdot\rVert_1$ is a Ky Fan norm and $\lVert\cdot\rVert_2$ is some other norm).

The conclusion holds for any vector norm.
If $\lVert x + y\rVert = \lVert x\rVert + \lVert y \rVert$ then
for $0 \le t \le 1$:
$$
\lVert x\rVert + \lVert y \rVert  = \lVert x + y\rVert \\
 = \lVert (1-t)x + tx + (1-t)y + ty\rVert \\
 \le (1-t)\lVert x \rVert +  \lVert  tx + (1-t)y \rVert + t \lVert y\Vert \\
 \stackrel{(*)}{\le} (1-t)\lVert x \rVert +  t \lVert  x \rVert + (1-t)\lVert y \rVert + t\lVert y\rVert \\
= \lVert x\rVert + \lVert y \rVert 
$$
so that equality holds everywhere, in particular at $(*)$: 
$$
\lVert tx + (1-t)y\rVert = t\lVert x\rVert + (1-t)\lVert y \rVert
$$
It follows even that for all $s, t > 0$
$$
\lVert sx + ty\rVert = s\lVert x\rVert + t\lVert y \rVert \, ,
$$
as can be seen by replacing $t$ by $\frac{s}{s+t}$ in the previous identity.
