Consider an entangled bipartite quantum state $\rho \in \mathcal{M}_d(\mathbb{C}) \otimes \mathcal{M}_{d'}(\mathbb{C})$ which is positive under partial transposition, i.e., $\rho^\Gamma \geq 0$. As separability of $\rho$ is equivalent to separability of its partial transpose $\rho^\Gamma$, we know that $\rho^\Gamma$ is entangled. Does this imply that the sum $\rho + \rho^\Gamma$ (ignoring trace normalization) is also entangled? If not, can we impose restrictions on $\rho$ which guarantee that the above proposition holds?
In the language of entanglement witnesses, the problem reduces to finding a common witness that detects both $\rho$ and $\rho^\Gamma$. Let $W$ be the entanglement witness detecting $\rho$, i.e., $\text{Tr} (W\rho) < 0$. Then $W$ is non-decomposable (as $\rho$ is PPT) and is of the canonical form $P+Q^\Gamma - \epsilon \mathbb{I}$, where $P, Q \geq 0$ are such that $\text{range}(P) \subseteq\text{ker}(\delta)$ and $\text{range}(Q) \subseteq \text{ker}(\delta^\Gamma)$ for some bipartite edge state $\delta$ (these are special states that violate the range criterion for separability in an extreme manner, see edge states) and $0 < \epsilon \leq \text{inf}_{|e,f\rangle} \langle e,f | P+Q^\Gamma | e,f \rangle$. If $\delta$ is such that $\text{ker}(\delta) \cap \text{ker}(\delta^\Gamma)$ is not empty, then we can choose $P=Q$ to be the orthogonal projector on $\text{ker}(\delta) \cap \text{ker}(\delta^\Gamma)$, in which case $W=W^\Gamma$ is the common witness. But is this always true? Can we use optimization of entanglement witness to ensure this condition?
