Cassini oval is convex for sufficiently large product constant

Question

Question

Given an integer $n\ge2$ and points $P_1,P_2,\dots,P_n\in\mathbb R^2$, then the set$$\{P\in\mathbb R^2:|PP_{1}|\times|PP_{2}|\times\dots\times|PP_n|\le b^n\}$$ is convex for sufficiently large $b$.

Intuitively, as $b\to\infty$, the curve looks more and more like a circle, so it'll eventually be convex.

My proof for $n=2$:

According to Wikipedia, the polar equation of Cassini oval is $$r^{4}-2a^{2}r^{2}\cos(2\theta)=b^{4}-a^{4}.$$ I try to prove:

The curve is convex for $\frac ba\geq {\sqrt {2}}$.

Let $a=1$, the equation is $$\tag1\label1 r^{4}-2r^{2}\cos(2\theta)-b^4+1=0$$ solving for $r^2$ $$r^2 = \cos(2θ)\pm\sqrt{\cos^2(2θ)+b^4 - 1}$$ The two curves with positive sign and negative sign are symmetric wrt $θ=\frac\pi4$.
so wlog take positive sign: $$\tag2\label2 r^2=\cos(2θ)+\sqrt{\cos^2(2θ)+b^4 - 1}$$ From a post the curve is convex iff $r^2 + 2 (r')^2 - r r'' \ge 0$ for all $θ$, substituting in $r',r''$, $$ r^2 + 2 (r')^2 - r r''=\frac{r^2}{(r^2-\cos (2 θ))^3}(r^6-r^4 \cos (2 θ)-r^2 \cos (4 θ)+\cos (2 θ)) $$ since $r^2-\cos (2 θ)>0$, we need to prove $$ r^6-r^4 \cos (2 θ)-r^2 \cos (4 θ)+\cos (2 θ)\ge0 $$ since $\eqref{1}=0$, subtracting $(r^2+\cos (2 θ))\times\eqref{1}$, we get $$ b^4 \left(r^2+\cos (2 θ)\right)\ge0 $$ equivalent to $$ r^2+\cos (2 θ)\ge0 $$ substituting $r^2$ from \eqref{2} $$ \forallθ:\quad2\cos(2θ)+\sqrt{\cos^2(2θ)+b^4 - 1}\ge0 $$ which is equivalent to $b\ge\sqrt2$.

QED.

---

How to prove the question for $n>2$?

Answer 1

The following argument feels too elementary - I feel like I might be missing some subtle issue but can't figure it out.

Anyways, let $d_i(x,y) = (x-a_i)^2 + (y-b_i)^2$ be the distance-squared function to the given point $P_i = (a_i,b_i)$ and consider the equation of our oval: $$0 = f(x,y) = \prod_{i=1}^n d_i(x,y) - b^{2n}.$$ Do an implicit differentiation to get $\dfrac{dy}{dx}:$ $$\dfrac{d}{dx}\left(d_i(x,y)\right) = 2(x-a_i) + 2(y-b_i)\dfrac{dy}{dx}$$ and so $$0=2d_1d_2\ldots d_n\sum_{i=1}^n\dfrac{(x-a_i) + (y-b_i)\tfrac{dy}{dx}}{d_i(x,y)}\implies \dfrac{dy}{dx} = -\dfrac{\sum_{i=1}^n\frac{x-a_i}{d_i}}{\sum_{i=1}^n\frac{y-b_i}{d_i}}.$$ The point is now that given $(x_0, y_0)$ that is on the boundary of our oval, we can calculate the tangent line $l(x)$ at that point: $$l(x) = \dfrac{dy}{dx}(x_0,y_0)(x-x_0) + y_0.$$ So it suffices to prove that for a sufficiently large $b>0$, the following system: \begin{equation} \dfrac{dy}{dx}(x_0,y_0)(x-x_0) + y_0 = y \\ f(x,y) = 0 \end{equation} only has the unique solution $(x,y) = (x_0,y_0).$ But note that given $(x_0,y_0),$ $y$ is a linear function of $x$, say $y = mx+n,$ from the first equation above. Then, the polynomial equation: $$f(x,mx+n) - b^{2n} = 0$$ is of degree $2n$ with leading coefficient $(1+m^2)^{n} > 0,$ therefore for a sufficiently large $b>0,$ it will only have a unique solution. The proof idea is simple - roughly speaking subtracting a large enough $b^{2n}$ will pull the graph downwards enough that it crosses the real axis only once because $\lim\limits_{x\to\infty}f(x,y) = \infty$.