Convexity of a certain sublevel set Consider the polynomial of degree $4$ in variable $r$
$$ r^4 + (x^2 + y^2)\ r^2 - 2 x y\ r + x^2 y^2 $$
The discriminant of this polynomial in $r$ is the following expression
(obtained using Mathematica)
$$ \Delta :=  16 x^2 y^2 \left( -x^6 + x^8 - 27 x^2 y^2 + 33 x^4 y^2 - 4 x^6 y^2 + 33 x^2 y^4 + 6 x^4 y^4 - y^6 - 4 x^2 y^6 + y^8 \right) $$
Consider the subset $D$ of $\mathbb{R}^2$ defined by the polynomial inequality $\frac{\Delta}{16 x^2 y^2} \le 0$
$$ D : -x^6 + x^8 - 27 x^2 y^2  - 4 x^6 y^2 - 4 x^2 y^6 + 33 x^4 y^2 +  33 x^2 y^4 + 6 x^4 y^4 - y^6  + y^8 \le 0 $$
I would like to prove that $D$ is convex. Any help would be appreciated. Thanks!
 A: This is more of an extended comment than an answer, but here goes.
It might be easier to consider the problem locally and then argue how global convexity follows. Local considerations don't tell if $D$ is connected. Let $f(x,y)=\Delta(x,y)/16x^2y^2$.
Let $z=(x,y)\in\partial D$ and let $\nu(z)$ be the unit outer normal at $z$. Take any vector $a$ orthogonal to $\nu(z)$. We would like to show that $Q(a)=\langle a,\frac{d}{dt}\nu(z+ta)|_{t=0}\rangle\geq0$. (Draw a figure to see why this means convexity near $z$. Or compare with the definition of the second fundamental form.)
The unit outer normal at $z$ is $\nabla f(z)/|\nabla f(z)|$. Using $\frac{d}{dt}\nabla f(z+ta)|_{t=0}=D^2f(z)a$ we get
$$
\frac{d}{dt}\nu(z+ta)|_{t=0}=|\nabla f|^{-3}(|\nabla f|^2D^2 fa-\langle\nabla f,D^2f a\rangle\nabla f).
$$
Thus
$$
Q(a)
=
|\nabla\Delta|^{-3}a^T A a,
$$
where $A=A(z)$ is the matrix
$$
A=\langle\nabla f,\nabla f\rangle D^2 f-(D^2 f)(\nabla f)(\nabla f)^T.
$$
It remains to check that $A(z)$ is positive definite for all $z\in\partial D$.
This seems tedious, but might well be doable.
A: You can parametrize the zero locus of $\Delta$ (other than the origin) in the first quadrant by
$$
(x,y) = \left(\frac{(3{-}t)\sqrt{(3{+}t)(1{-}t)}}{8},
              \frac{(3{+}t)\sqrt{(3{-}t)(1{+}t)}}{8}\right)
   \qquad\qquad -1\le t\le 1
$$
and then compute that the curvature is 
$$
\frac{(x'y''-y'x'')}{((x')^2+(y')^2)^{3/2}}
 = \frac{2^{3/2}(3{+}t^2)}{\bigl(3{-}t^2\bigr)^{5/2}}.
$$
Since this is always positive, it follows that the curve is strictly convex (since it is symmetric under $x$ and $y$ reflections.
