\subsection{A bound for each degree}\label{sec:DEG}

In the proof of~\cite[Thm.~7.1]{JatarNg2026b} the inverse branch of $F$ fixing $0$ omits every zero of $B$, and Pick's estimate at one
zero gives $\mathcal B_d<4-\frac{529}{576}\frac{(\log2)^2}{d^2}$~\cite[Thm.~1.3, Cor.~8.10]{JatarNg2026b}; \cite[Rem.~7.3]{JatarNg2026b}
suggests that the hyperbolic geometry of $\mathbb D$ minus all the zeros does better. We confirm this for large $d$.
Here we use Lemma~\ref{lem:P}: the map $h$ of its proof is univalent and omits all of $A$. By~\cite[Lemma~2.1]{JatarNg2026b},
$\mathcal B_d$ is also the supremum of $K_0$ over the Blaschke models with $\deg B\le d$: regular admissible pairs are Blaschke models, the regularisation
keeps $\deg B$, and $B\equiv1$ gives $1/B(0)=1$.

We use the following constants. Fix
\[
  \delta_0=10^{-3},\quad \epsilon_{\max}=\sqrt{3\delta_0/2},\quad \ell_{\max}=\log\frac1{1-\epsilon_{\max}},\quad
  c_2=\frac{(2-\epsilon_{\max})^2(1-\ell_{\max}/2)^2}{1+4\pi^2},
\]
\[
  c_3=\frac4{\log(1+c_2)},\quad c_0=1+c_3\log\frac1{1-\ell_{\max}/2},\quad c_1=2\log\cosh\frac12,\quad
  {\hat M}_0=\log(4-\delta_0),
\]
\[
  q_d=c_3\log\frac{4d}{\hat M_0}+c_0,\qquad Y_d=\frac{c_1\hat M_0^2}{4d\,q_d}.
\]
Here $c_2>0.091311$, $c_3<45.778$, $c_0<1.9132$, $c_1>0.240229$ and $\hat M_0>1.386044$. The constants enter as follows:
$\delta_0$ is the threshold $K_0\ge4-\delta_0$ above which Lemma~\ref{lem:select} is used; $\epsilon_{\max}$ and $\ell_{\max}$ bound
$1-|a|$ and $\log(1/|a|)$ for the zeros $a$ of such a Blaschke model (Corollary~\ref{cor:P}(c)); $c_2$, $c_3$ and $c_0$ enter the bound
$\operatorname{mass}(a)\le\ell_a\,q(\ell_a)$ for the mass of a group of zeros in the proof of Lemma~\ref{lem:select}, with
$\ell_a=\log(1/|a|)$ and $q(\ell)=c_3\log(2/\ell)+c_0$, and $q_d=q(\hat M_0/(2d))$;
$c_1=2\log\cosh\frac12$ comes from Lemma~\ref{lem:sector} with $y=\frac12$; and $\hat M_0\le\log K_0=\sum_{a\in A}\log(1/|a|)$.

\begin{theorem}\label{thm:DEG}
For every $d\ge2$, $\mathcal B_d\le4-5.97\cdot10^{-3}/(d\log d)$, and
$\liminf_{d\to\infty}(4-\mathcal B_d)\,d\log d\ge1.005\cdot10^{-2}$. More precisely, $\mathcal B_d\le4-3.99\,Y_d$ for every $d\ge1$.
\end{theorem}

The bound $4-5.97\cdot10^{-3}/(d\log d)$ is stronger than~\cite[Thm.~1.3]{JatarNg2026b} for every
$d\ge452$, and the bound $4-3.99\,Y_d$ for every $d\ge299$.

Proposition~\ref{prop:sep} and Lemma~\ref{lem:sector} below follow Dubinin's proof of~\cite[Thm.~2.1]{Dubinin1990}: his separating
transformation, with the maps $\Xi\circ p_j$, $\Xi(z)=z/(1+z^2)$ as in the proof of Lemma~\ref{lem:sector} (the map $p_j$ of
Proposition~\ref{prop:sep} sends the bisector of $\mathcal C_j$ to the positive axis), and then a Koebe step. Since we need sectors of different
openings, we include the proofs. The bounds
$K(n)\le4^{(n-2)/(n-1)}$~\cite{BeardonMindaNg2002}, $K(n)\le4(n-1)/(n+1)$~\cite{ConteFujikawaLakic2007} and
$K(n)<4-2.263/\sqrt n$ for $n\ge8$~\cite{Crane2007} of Section~\ref{sec:intro} concern polynomials of degree $n$; they do not bound $\mathcal B_d$, since
the pairs with $\deg B\le d$ are not limited to pairs that come from polynomials of bounded degree.

For a domain $U\ni0$ in the Riemann sphere with a classical Green function $g_U$ let $r(U,0)$ be its inner radius at $0$:
$g_U(z,0)=\log(1/|z|)+\log r(U,0)+o(1)$ as $z\to0$. For simply connected $U$ this is the conformal radius, also when $\infty\in U$.
We have $r(U,0)\le r(U',0)$ if $U\subset U'$, and $r$ does not change under a conformal map that fixes $0$ with derivative $1$ there.

\begin{proposition}[after Dubinin~{\cite[Lemma~1.2]{Dubinin1990}}]\label{prop:sep}
Let $U$ be a domain with $0\in U\subset\mathbb D$ and a classical Green function. Let $n\ge2$, $\theta_1<\dots<\theta_n<\theta_1+2\pi$, let
$\mathcal C_j=\{\theta_j<\arg w<\theta_{j+1}\}$ \textup{(}$\theta_{n+1}=\theta_1+2\pi$\textup{)} have opening $\pi\alpha_j$, so that $0<\alpha_j<2$ and
$\sum_j\alpha_j=2$, and let $p_j(w)=-i(e^{-i\theta_j}w)^{1/\alpha_j}$, which maps $\mathcal C_j\cap\mathbb D$ conformally onto the right half-disk
and extends continuously and injectively to the sides. Let $\Omega_j$ be the component containing $0$ of the
open set
\[
  O_j=p_j(U\cap {\mathcal C}_j)\cup\bigl\{-\bar\zeta:\zeta\in p_j(U\cap {\mathcal C}_j)\bigr\}\cup\{p_j(w):w\in U\cap\partial {\mathcal C}_j,\ w\neq0\}\cup\{0\}.
\]
Then $\log(1/r(U,0))\ge\frac12\sum_j\alpha_j^2\log(1/r(\Omega_j,0))$.
\end{proposition}

\begin{proof}
For a domain $V\ni0$ with a classical Green function and small $\rho>0$, let $\operatorname{cap}(\rho,V)$ be the infimum of the Dirichlet
integrals $\int|\nabla v|^2$ over the Lipschitz functions $v$ with compact support in $V$ and $v=1$ on $\{|z|\le\rho\}$. Then
\[
  \operatorname{cap}(\rho,V)=\frac{2\pi}{\log(1/\rho)+\log r(V,0)+o(1)}\qquad(\rho\to0).
\]
Indeed, for $t>0$ the function $\min(g_V(\cdot,0)/t,1)$ has Dirichlet integral $2\pi/t$, the least one for the level set
$\{g_V\ge t\}$ in place of the disk; the capacity increases with the set; and for small $\rho$ the disk $\{|z|\le\rho\}$ lies between the
level sets with $t=\log(1/\rho)+\log r(V,0)\pm o(1)$.

Let $\epsilon>0$ be small, $\hat L=\log(1/\epsilon)$ and $\eta>0$, and let $u$ be a Lipschitz function with compact support in $U$, $u=1$ on
$\{|w|\le\epsilon\}$, and $\int|\nabla u|^2\le\operatorname{cap}(\epsilon,U)+\eta$. The sides of the sectors have measure zero, so
$\int|\nabla u|^2=\sum_j\int_{U\cap {\mathcal C}_j}|\nabla u|^2$. Fix $j$, and define $u_j$ on $O_j$ by $u_j=u\circ p_j^{-1}$ on
$p_j(U\cap {\mathcal C}_j)$ and on the image of $U\cap\partial {\mathcal C}_j\setminus\{0\}$, by $u_j(-\bar\zeta)=u_j(\zeta)$, and by $u_j(0)=1$. Away from $0$ the map
$p_j$ extends conformally across each side of $\mathcal C_j$ and maps it into the imaginary axis, which the reflection $\zeta\mapsto-\bar\zeta$ fixes; so $O_j$ is open and $u_j$
is Lipschitz. As $u=1$ near $0$ and $|p_j(w)|=|w|^{1/\alpha_j}$, $u_j$ has compact support in $O_j$ and $u_j=1$ on
$\{|\zeta|\le\epsilon^{1/\alpha_j}\}$. The Dirichlet integral is invariant under the conformal map $p_j$ and under $\zeta\mapsto-\bar\zeta$, so
\[
  \int_{O_j}|\nabla u_j|^2=2\int_{U\cap {\mathcal C}_j}|\nabla u|^2 .
\]
Since $\Omega_j$ is open and closed in $O_j$, the restriction of $u_j$ to $\Omega_j$, extended by $0$, competes in
$\operatorname{cap}(\epsilon^{1/\alpha_j},\Omega_j)$, and $\Omega_j\subset\mathbb D$ has a classical Green function. Hence
\[
  \operatorname{cap}(\epsilon,U)+\eta\ \ge\ \frac12\sum_j\operatorname{cap}(\epsilon^{1/\alpha_j},\Omega_j).
\]
Let $\eta\to0$ and use the asymptotics above, with $\log(1/\epsilon^{1/\alpha_j})=\hat L/\alpha_j$:
\[
  \frac{2\pi}{\hat L+\log r(U,0)+o(1)}\ \ge\ \sum_j\frac{\pi\alpha_j}{\hat L+\alpha_j\log r(\Omega_j,0)+o(1)}\qquad(\hat L\to\infty).
\]
Since $1/(\hat L+b+o(1))=1/\hat L-b/\hat L^2+o(1/\hat L^2)$ for fixed $b$, and $\sum_j\alpha_j=2$, both sides equal $2\pi/\hat L$ plus terms of order $1/\hat L^2$:
\[
  \frac{2\pi}{\hat L}-\frac{2\pi\log r(U,0)}{\hat L^2}+o\Bigl(\frac1{\hat L^2}\Bigr)\ \ge\ \frac{2\pi}{\hat L}-\frac{\pi\sum_j\alpha_j^2\log r(\Omega_j,0)}{\hat L^2}
  +o\Bigl(\frac1{\hat L^2}\Bigr).
\]
Multiplying by $\hat L^2/(2\pi)$ and letting $\hat L\to\infty$ gives $\log(1/r(U,0))\ge\frac12\sum_j\alpha_j^2\log(1/r(\Omega_j,0))$.
Dubinin~\cite{Dubinin1990} proves the inequality with level sets of $g_U$; the transformation was introduced in~\cite{Dubinin1986}
(see~\cite[p.~54]{Dubinin1990}).
\end{proof}

\begin{lemma}[one sector]\label{lem:sector}
Let $U$ be a simply connected domain with $0\in U\subset\mathbb D$, let $a\in\mathbb D\setminus U$, $a\neq0$, and $0<\alpha<2$. Apply
Proposition~\ref{prop:sep} to $U$ and a sector $\mathcal C$ of opening $\pi\alpha$ whose bisector passes through $a$, and let $\Omega$ be the
corresponding domain. Then $\log(1/r(\Omega,0))\ge\log\cosh y$, where $y=\log(1/|a|)/\alpha$, with equality when $U$ is $\mathbb D$
minus the radial slit from $a$ to the circle.
\end{lemma}

\begin{proof}
This is the Koebe step of Dubinin's proof of~\cite[Thm.~2.1]{Dubinin1990}, in our normalization. Here $p(w)=(w|a|/a)^{1/\alpha}$ and
$p(a)=x=e^{-y}$. The set $X=\widehat{\mathbb C}\setminus U$ is a continuum, and $E=X\cap {\mathcal C}\cap\mathbb D$ is a proper open subset of $X$
containing $a$. By the boundary bumping theorem~\cite[Thm.~5.6]{Nadler1992}, applied to $X\setminus E$, the closure of the component of
$E$ containing $a$ meets a side of $\mathcal C$ or the circle; let $K'$ be its image under $p$. This is a continuum in the closed right half-disk,
it contains $x$ and a point of the circle or of the imaginary axis, and $K'\cap\mathbb D$ and $\{-\bar z:z\in K'\}\cap\mathbb D$ miss $\Omega$.

The map $\Xi(z)=z/(1+z^2)=1/\mathcal J(z)$ sends $\mathbb D$ conformally onto $\mathbb C\setminus\Sigma$, $\Sigma=(-\infty,-\frac12]\cup[\frac12,\infty)$,
with $\Xi(0)=0$, $\Xi'(0)=1$ and $\Xi(-\bar z)=-\overline{\Xi(z)}$; it extends continuously to a map of the closed disk into $\widehat{\mathbb C}$ that sends
the circle onto $\Sigma\cup\{\infty\}$ and the imaginary axis into itself. So $r(\Xi(\Omega),0)=r(\Omega,0)$, and the continua $\Xi(K')$ and
$\{-\bar z:z\in\Xi(K')\}$ miss $\Xi(\Omega)$ and contain $\zeta=\Xi(x)=1/(2\cosh y)$ and $-\zeta$. Either both meet the connected set $\Sigma\cup\{\infty\}$,
or they share a point of the imaginary axis. Hence $\pm\zeta$ lie in one component $E'$ of $\widehat{\mathbb C}\setminus \Xi(\Omega)$. The
component $V$ of $\widehat{\mathbb C}\setminus E'$ containing $0$ is simply connected, contains $\Xi(\Omega)$ and omits $\pm\zeta$. Let $f$
map $\mathbb D$ conformally onto $V$ with $f(0)=0$, $f'(0)>0$. Then $f_1=\zeta f/(\zeta-f)$ is holomorphic and univalent, $f_1(0)=0$,
$f_1'(0)=f'(0)$, and $f_1$ omits $-\zeta/2$; by Koebe's one-quarter theorem $f'(0)\le2\zeta$. So $r(\Omega,0)\le r(V,0)\le2\zeta=1/\cosh y$.
For the slit domain, $\Omega$ is $\mathbb D$ minus two symmetric slits and $r(\Omega,0)=2x/(1+x^2)=1/\cosh y$.
\end{proof}

\begin{lemma}[selection]\label{lem:select}
Let $(W,B)$ be a Blaschke model with $k=\deg B$, and let $K_0\ge4-\delta_0$. For $a\in A$ put $\ell_a=\log(1/|a|)$. Then there is
$A_0\subset A$ such that the arcs $\{e^{it}:|t-\arg a|<\pi\ell_a\}$, $a\in {A_0}$, are pairwise disjoint and $\sum_{a\in {A_0}}\ell_a^2\ge
{\hat M}_0^2/(4kq_k)$.
\end{lemma}

\begin{proof}
By Corollary~\ref{cor:P}(c), $\epsilon_a=1-|a|\le\epsilon_{\max}$ and $\ell_a\le\ell_{\max}<1/2$ for every $a\in A$, and
$\sum_{a\in A}\ell_a=\log K_0\ge {\hat M}_0$. Order $A$ by decreasing $\ell_a$. Take $a$ into $A_0$ if its arc misses the arcs already taken. Assign
each remaining $b$ to an earlier $a\in {A_0}$ whose arc meets that of $b$. Then $|b|\ge|a|$, and the angle $t$ between $a$ and $b$ is
less than $2\pi\ell_a$, so $|a-b|^2\le\epsilon_a^2+t^2\le\ell_a^2(1+4\pi^2)$. As $\epsilon\ge\ell-\ell^2/2$, we have
$1-|a|^2=\epsilon_a(2-\epsilon_a)\ge(2-\epsilon_{\max})(1-\ell_{\max}/2)\,\ell_a$, and the same for $b$; so
\[
  \frac{|1-\bar ab|^2}{|a-b|^2}=1+\frac{(1-|a|^2)(1-|b|^2)}{|a-b|^2}\ \ge\ 1+c_2\,\frac{\ell_b}{\ell_a}.
\]
Hence $g_{\mathbb D}(a,b)=\log(|1-\bar ab|/|a-b|)\ge\frac12\log(1+c_2\ell_b/\ell_a)\ge\frac{\ell_b}{2\ell_a}\log(1+c_2)$, the last step by
concavity, as $\ell_b\le\ell_a$. By Lemma~\ref{lem:Z} and Koebe's growth bound, $\sum_{b\neq a}g_{\mathbb D}(a,b)\le\log(4/\epsilon_a^2)$.
So the group of $a$ ($n_a$ points, $a$ included; $\operatorname{mass}(a)$ is the sum of their $\ell_b$) satisfies, as $\epsilon_a\ge(1-\ell_{\max}/2)\,\ell_a$,
\[
  \operatorname{mass}(a)-\ell_a\ \le\ \frac{2\ell_a}{\log(1+c_2)}\log\frac4{\epsilon_a^2}=c_3\ell_a\log\frac2{\epsilon_a}
  \ \le\ c_3\ell_a\Bigl(\log\frac2{\ell_a}+\log\frac1{1-\ell_{\max}/2}\Bigr);
\]
that is, $\operatorname{mass}(a)\le\ell_a\,q(\ell_a)$ with $q(\ell)=c_3\log(2/\ell)+c_0$. Also $\operatorname{mass}(a)\le n_a\ell_a$. Let $\eta=\hat M_0/(2k)$, so that $q(\eta)=q_k$.
The groups with $\ell_a<\eta$ carry less than $k\eta=\hat M_0/2$; for the others, $\operatorname{mass}(a)^2\le n_a\ell_a^2q_k$. By the Cauchy--Schwarz
inequality and $\sum n_a\le k$, $(\hat M_0/2)^2\le k\,q_k\sum_{a\in {A_0}}\ell_a^2$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:DEG}]
For $d=1$, $\mathcal B_1<\sqrt{7/5}<4-3.99Y_1$ by~\cite[Cor.~8.10]{JatarNg2026b}. For $d=2,3,4$, \cite[Thm.~1.3]{JatarNg2026b} gives
$\mathcal B_d<4-\frac{529}{576}(\log2)^2/d^2\le4-3.99Y_d$ (at $d=4$: $\frac{529}{576}(\log2)^2/16>0.0275$ and $3.99Y_4<0.0011$). Let $d\ge5$ and let $(W,B)$ be a Blaschke model with
$k=\deg B\le d$. If $K_0<4-\delta_0$, then $K_0<4-3.99Y_d$, since $3.99Y_5\le\delta_0$ and $Y_d\le Y_5$. Otherwise take $A_0$ from
Lemma~\ref{lem:select}. By the proof of Lemma~\ref{lem:P}, $U=h(\mathbb D)$ is a simply connected domain with $0\in U\subset
\mathbb D\setminus A$ and $r(U,0)=r_{\rm in}K_0\ge K_0/4$. Give each $a\in {A_0}$ the sector with bisector through $a$ and opening $2\pi\ell_a$,
that is $\alpha_a=2\ell_a<1$. These sectors are disjoint, so their total opening is at most $2\pi$. If it equals $2\pi$, they are all
the sectors, and $n=|A_0|\ge2$. Otherwise the rest of the circle is a finite union of closed arcs (possibly single points), and we add one
sector over the interior of each non-degenerate one; again $n\ge2$ and every $\alpha<2$. Proposition~\ref{prop:sep} and Lemma~\ref{lem:sector} with $y=1/2$
give
\[
  \log\frac4{K_0}\ge\log\frac1{r(U,0)}\ge\frac12\sum_{a\in {A_0}}4\ell_a^2\log\cosh\frac12
  =c_1\sum_{a\in {A_0}}\ell_a^2\ge\frac{c_1\hat M_0^2}{4kq_k}=Y_k\ge Y_d.
\]
As $Y_d\le Y_5<0.0002$, $K_0\le4e^{-Y_d}\le4-3.99Y_d$. For the consequences, $Y_d\,d\log d$ increases in $d$ and tends to $c_1\hat M_0^2/(4c_3)$,
and $Y_dd^2$ increases for $d\ge2$. A direct evaluation gives $3.99\,Y_5\cdot5\log5=5.97005\ldots\cdot10^{-3}>5.970\cdot10^{-3}$, while for $d=2,3,4$ the bound
of~\cite[Thm.~1.3]{JatarNg2026b} times $d\log d$ exceeds $0.15$; further $3.99\,c_1\hat M_0^2/(4c_3)>1.0056\cdot10^{-2}$ and
$3.99\,Y_{299}>4.9446\cdot10^{-6}>4.9357\cdot10^{-6}>\frac{529}{576}(\log2)^2/299^2$; also
$5.97\cdot10^{-3}/(452\log452)>2.1603\cdot10^{-6}>2.1598\cdot10^{-6}>\frac{529}{576}(\log2)^2/452^2$, and $d/\log d$ increases for $d\ge3$.
\end{proof}

\begin{remark}
For a single radial slit from $a$ to the circle equality holds in Lemma~\ref{lem:sector}, and Proposition~\ref{prop:sep} gives $0.2402\,\ell_a^2$
for one slit, against the exact $2\log\cosh(\ell_a/2)\sim\ell_a^2/4$. For $d$ points equally spaced on $|z|=r$ with $r^d=1/4$ (these points cannot be the
zeros of $B$ in a Blaschke model, since then $K_0=4$; the example tests only Lemma~\ref{lem:select}), Dubinin's theorem~\cite[Thm.~3.2]{Dubinin1990} gives
$\log(1/r(U,0))\ge\frac2d\log\frac54$ for every simply connected $U\ni0$ in $\mathbb D$ that misses these points, with equality for the
symmetric slit domain; the selection of Lemma~\ref{lem:select}, taking the zeros in their order on the circle, keeps every second zero, and the
bound $c_1\sum_{a\in A_0}\ell_a^2$ of the proof of Theorem~\ref{thm:DEG} is then about $0.230/d$. The remaining
$\log d$ comes from Lemma~\ref{lem:Z}, which allows groups of about $\log(1/\ell_a)$ zeros at mutual distance $\approx\ell_a$. Whether
$4-\mathcal B_d\ge c/d$ holds is open.
\end{remark}