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\textbf{\textit{Proof of property 1.}}
\bigskip{}
For all $b$ and $k$, let $\phi\left(b,k\right)$ be defined by $\phi\left(b,k\right)\equiv8\left(b-k\right)^{2}+\left(1-k\right)\left(1-2b+k^{2}\right)$.
\bigskip{}
...
\begin{itemize}
\item If $\delta\leq k$, we show that
\end{itemize}
$$
{\textstyle \frac{\partial}{\partial k}\left(x^{*}\left(\delta\right)+y^{*}\left(\delta\right)\right){\textstyle =-\frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\sqrt{\frac{b-\delta}{b-k}}\frac{\phi\left(b,k\right)}{2}},}
$$
...
-x^{*}\left(\delta\right)-y^{*}\left(\delta\right)
\end{array}\right)}{\textstyle =\frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\sqrt{\frac{b-\delta}{b-k}}\phi\left(b,k\right).}
$$
\end{itemize}
\bigskip{}
\begin{itemize}
\item If $\delta>k$, we show that
\end{itemize}
$$
{\textstyle \frac{\partial}{\partial k}\left(x^{*}\left(\delta\right)+y^{*}\left(\delta\right)\right){\textstyle =-\frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\left(4\left(b-k\right)^{2}+\left(1-k\right)^{3}\right)},}
$$
...
\end{array}\right)}.}
$$
\end{itemize}
\bigskip{}
Property 1 directly follows, assuming that $b$ and $k$ are such that $\phi\left(b,k\right)>0$. To obtain the last sign, denote $\varphi\left(b,k,\delta\right)\equiv\phi\left(b,k\right)+2\left(1-2b+k\right)\left(\delta-k\right)$.
Then:
\begin{itemize}
\item If $1-2b+k\geq0$, then $\varphi\left(b,k,\delta\right)>0$ directly follows from $\delta>k$ and $\phi\left(b,k\right)>0$;
\item If $1-2b+k<0$, then $\varphi\left(b,k,\delta\right)$ is decreasing in $\delta$ and is bounded below by $\varphi\left(b,k,1\right)$.
\end{itemize}
Then remark that $\varphi\left(b,k,1\right)=8b^{2}-2\left(3+5k\right)b+\left(3-k+7k^{2}-k^{3}\right)$ and that this quadratic polynomial in $b$ is strictly positive.\footnote{Because the discriminant $\Delta=-4(15-8k)(1-k)\text{\texttwosuperior}<0$ and $\varphi\left(k,k,1\right)=\left(3-k\right)\left(1-k\right)^{2}>0$.}
\bigskip{}
Finally, we can check that the condition that $\phi\left(b,k\right)>0$ is not very restrictive. Rearrange $\phi\left(b,k\right)$ to write it as $\phi\left(b,k\right)=8b^{2}-2\left(1+7k\right)b+\left(1-k+9k^{2}-k^{3}\right)$.
We have:
\begin{itemize}
\item If $k<7/8$, this quadratic polynomial in $b$ is strictly positive;\footnote{Because the discriminant $\text{\textgreek{D}}=4\left(8k-7\right)\left(1-k\right)^{2}<0$ and $\phi\left(k,k\right)=\left(1-k\right)^{3}>0$.}
\item If $k\geq7/8$, it is positive if $b\notin\left[k+\left(1-k\right)\left(1-\sqrt{8k-7}\right)/8,k+\left(1-k\right)\left(1+\sqrt{8k-7}\right)/8\right]$.
\end{itemize}
$\blacksquare$
