Why does entanglement negativity not satisfy the triangle inequality in the usual sense? I am a bit puzzled, I’ve read in some places, like the original paper by Vidal, that
$$
\mathcal{N}(\sum_n a_n \rho_n)\leq \sum_n a_n \mathcal{N}(\rho_n)
$$
whenever $a_n \geq 0$ and $\sum_n a_n =1$. I am guessing the more general inequality holds too:
$$
\mathcal{N}(\sum_n a_n \rho_n)\leq \sum_n |a_n| \mathcal{N}(\rho_n), \tag{1}
$$
for any $a_n\in\mathbb{C}$, not just $a_n\geq 0$. This should follow from the standard operator norm properties and the original paper was likely just thinking about a general mixed state than a general operator.
Am I missing something? Is expression (1) correct?
NB: $\mathcal{N}(\rho) = (||\rho^{T_A}||_1-1)/2$, with $||\cdot||_1$ the trace norm.

Edit: As pointed out by user glS expression (1) holds provided that $\sum_n |a_n| \leq 1$.
 A: Let $a_n\in\mathbb C$ be generic complex coefficients. The convexity and (positive) homogeneity of the trace norm then tells you that
$$\mathcal N(\rho)
= \mathcal N(\sum_n a_n \rho_n)
\equiv \frac12 \left(\bigg\|\sum_n a_n \rho_n^{T_A}\bigg\|_1-1\right)
\le \frac12 \left(\sum_n |a_n|\, \|\rho_n^{T_A}\|_1-1\right).$$
Therefore if $a_n\ge0$ and $\sum_n a_n=1$, then you can conclude that
$$\mathcal N(\rho) \le \sum_n a_n \mathcal N(\rho_n),$$
but this is not true in general.
As a trivial counterexample, take any separable state $\rho$, e.g. $\rho=|00\rangle\!\langle00|$, and $a\in\mathbb C$. Then
$$\mathcal N(a\rho)=\frac{\|a\rho\|_1-1}{2}=\frac{|a|-1}{2},
\qquad
\mathcal |a| N(\rho) = |a|\frac{\|\rho\|_1-1}{2} = 0,$$
and thus whenever $|a|>1$ you have $\mathcal N(a\rho)>|a|\mathcal N(\rho)=0$.
A: $a_n \geq 0$ and $\sum_n a_n = 1$ is a condition making $\sum_n a_n \rho_n$ a so-called convex combination and a little bit stronger than what you would need. You can derive (1) using $\sum_n \left|a_n\right| = 1$ and properties that any norm has by definition.
