Is the set of two-qubit absolutely separable states convex?

Companion question on MathOverflow

Let us order the four nonnegative eigenvalues, summing to 1, of a 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 left(lambda _1-lambda _2right){}^2 left(lambda _1-lambda _3right){}^2 left(lambda
_2-lambda _3right){}^2 left(2 lambda _1+lambda _2+lambda _3-1right){}^2 left(lambda _1+2
lambda _2+lambda _3-1right){}^2 left(lambda _1+lambda _2+2 lambda _3-1right){}^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.

ThreeSetPlot