<s>[INST] 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!
 [/INST] 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.
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