Let $\mathcal{H},\mathcal{K}$ be Hilbert spaces ($\cong \mathbb{C}^2$ if necessary) and let $h_1,h_2$ denote semi-positive $\ge 0$ operators on $\mathcal{H}\otimes \mathcal{K}$. If $h_1,h_2$ are entangled, i.e., not a trivial product operator, then is this also true for $h_1 +h_2$? (not including the trivial case where $h_1+h_2=0$)
Update: Let me also assume that $h_1+h_2$ has a nontrivial kernel or $h_1+h_2$ is not of full rank.
Background. Physically, this corresponds to a system with nontrivial ground state and entangled excited state.
Thoughts. Whether $h$ is entangled or not is characterized by its range, i.e., whether it's range is equal to $V_1\otimes V_2$ with $\dim V_1 =1$ (or $\dim V_2=1$). Hence, the statement then boils to understanding how the range of $h_1+h_2$ is related to that of $h_1$ and $h_2$. This is where I'm stumped.
In $\mathbb{C}^2 \otimes \mathbb{C}^2$, you actually do get that when $h_1 + h_2$ is not of full rank, then it must be entangled. Indeed, if it is not entangled, then its range is of the form $V_1 \otimes V_2$. (The converse is false, as in your previous question, but this direction can be easily proved.) If it is not of full rank, then either $V_1$ or $V_2$ has to be one-dimensional. But then $h_1 + h_2 \leq Cp_1 \otimes p_2$ for some constant $C > 0$, where $p_i$ is the projection onto $V_i$. This implies $h_1 \leq Cp_1 \otimes p_2$, so $h_1$ has range in $V_1 \otimes V_2$. Since at least one of $V_1$ and $V_2$ is one-dimensional, as I mentioned in comments below my answer to your previous question (https://math.stackexchange.com/a/4885346/465145), this actually implies $h_1$ is a simple tensor, and the same holds for $h_2$.
Now, of course, when the dimension goes above $2$, $h_1 + h_2$ can be neither entangled nor of full rank. Just use the example in Severin's comment for dimension $2$ and enlarge one of the spaces.