Let us order the four nonnegative eigenvalues, summing to 1, of a (by definition, $4 \times 4$, Hermitian, nonnegative definite, trace one) "two-qubit density matrix" ($\rho$) as \begin{equation} 1 \geq x \geq y \geq z \geq (1-x-y-z) \geq 0. \end{equation} The set ($S$) of absolutely separable states (those that can not be "entangled" by global unitary transformations) is defined by the additional inequality (eq. (1) in Halder) \begin{equation} x - z \leq 2 \sqrt{y (1-x-y-z)}. \end{equation}

Is the set $S$, that is, \begin{equation} 1 \geq x \geq y \geq z \geq (1-x-y-z) \geq 0 \land x - z \leq 2 \sqrt{y (1-x-y-z)}, \end{equation} convex?

If so, I would like to seek to determine the John ellipsoids JohnEllipoids containing and contained within $S$ and see if they are simply the same as the circumscribed ($\mbox{Tr}(\rho^2) \leq \frac{3}{8}$) and inscribed ($\mbox{Tr}(\rho^2) \leq \frac{1}{3}$) sets, respectively Adhikari .

These two sets are determined by the constraints \begin{equation} 1 \geq x \geq y \geq z \geq (1-x-y-z) \geq 0 \land x^2 +y^2 +z^2 +(1-x-y-z)^2 \leq \frac{3}{8}. \end{equation} and \begin{equation} 1 \geq x \geq y \geq z \geq (1-x-y-z) \geq 0 \land x^2 +y^2 +z^2 +(1-x-y-z)^2 \leq \frac{1}{3}. \end{equation} (The latter set corresponds to the separable "maximal ball" inscribed in the set of two-qubit states (sec. 16.7 GeometryQuantumStates).)

Further, I am interested in the Hilbert-Schmidt probabilities (relative volumes) Hilbert-Schmidt of these various sets. These probabilities are obtained by integrating over these sets the expression \begin{equation} 9081072000 \Pi_{j<k}^4 (\lambda_j-\lambda_k)^2, \end{equation} where the four eigenvalues are indicated. (This integrates to 1, when only the eigenvalue-ordering constraint--given at the very outset--is imposed.)

In the answer to 4-ball, we report formulas for the Hilbert-Schmidt probabilities (relative volumes) of these inscribed and circumscribed sets, that is, \begin{equation} \frac{35 \pi }{23328 \sqrt{3}} \approx 0.00272132 \end{equation} and the considerably larger \begin{equation} \frac{35 \sqrt{\frac{1}{3} \left(2692167889921345-919847607929856 \sqrt{6}\right)} \pi}{27518828544} \approx 0.0483353. \end{equation} (We also have given an exact--but still quite cumbersome--formula [$\approx 0.00484591$] for $\mbox{Tr}(\rho^2) \leq \frac{17}{50}$.)

Further, in the answers to AbsSepVol1 and AbsSep2 , the formula for the Hilbert-Schmidt volume (confirming and rexpressing the one given in 2009paper) \begin{equation} \frac{29902415923}{497664}-\frac{50274109}{512 \sqrt{2}}-\frac{3072529845 \pi }{32768 \sqrt{2}}+\frac{1024176615 \cos ^{-1}\left(\frac{1}{3}\right)}{4096 \sqrt{2}} \approx 0.00365826 \end{equation} of the intermediate absolutely separable set $S$ has been given.

As to the total (absolute and non-absolute) separability probability of the 15-dimensional convex set of two-qubit density matrices, compelling evidence of various kinds--though yet no formalized proof--indicate that its value is the considerably larger $\frac{8}{33} \approx 0.242424$ MasterLovasAndai . (One can also enquire as to the John ellipsoids for this [known-to-be] convex set JohnEllipsoid2.)

Here is a joint plot of the three sets of central interest here.