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Proof of property 1.

For all \(b\) and \(k\), let \(\phi\left(b,k\right)\) be defined by \(\phi\left(b,k\right)\equiv 8\left(b-k\right)^{2}+\left(1-k\right)\left(1-2b+k^{2}\right)\).

We distinguish two cases:

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If \(\delta\leq k\), we show that

\begin{equation} {\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}},}\nonumber \\ \end{equation} \begin{equation} {\textstyle\frac{\partial}{\partial k}\pi\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\left(b-k\right)}},\nonumber \\ \end{equation} \begin{equation} {\textstyle\frac{\partial}{\partial k}\left(\begin{array}{c}\pi\left(x^{*}\left(\delta\right),y^{*}\left(\delta\right)\right)\left(b-\delta\right)\\ -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).}\nonumber \\ \end{equation}

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If \(\delta>k\), we show that

\begin{equation} {\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)},}\nonumber \\ \end{equation} \begin{equation} {\textstyle\frac{\partial}{\partial k}\pi\left(x^{*}\left(\delta\right),y^{*}\left(\delta\right)\right){\textstyle=-2\frac{1-k}{\left(2\left(b-k\right)+\left(1-k\right)^{2}\right)^{2}}\left(1-2b+k\right)},}\nonumber \\ \end{equation} \begin{equation} {\textstyle\frac{\partial}{\partial k}\left(\begin{array}{c}\pi\left(x^{*}\left(\delta\right),y^{*}\left(\delta\right)\right)\left(b-\delta\right)\\ -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}}\left(\begin{array}{c}\phi\left(b,k\right)\\ +2\left(1-2b+k\right)\left(\delta-k\right)\end{array}\right)}.}\nonumber \\ \end{equation}

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:

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If \(1-2b+k\geq 0\), then \(\varphi\left(b,k,\delta\right)>0\) directly follows from \(\delta>k\) and \(\phi\left(b,k\right)>0\);
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If \(1-2b+k<0\), then \(\varphi\left(b,k,\delta\right)\) is decreasing in \(\delta\) and is bounded below by \(\varphi\left(b,k,1\right)\).

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.¹¹Because the discriminant \(\Delta=-4(15-8k)(1-k)\text{\texttwosuperior}<0\) and \(\varphi\left(k,k,1\right)=\left(3-k\right)\left(1-k\right)^{2}>0\).

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:

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If \(k<7/8\), this quadratic polynomial in \(b\) is strictly positive;²²Because the discriminant \(\text{\textgreek{D}}=4\left(8k-7\right)\left(1-k\right)^{2}<0\) and \(\phi\left(k,k\right)=\left(1-k\right)^{3}>0\).
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If \(k\geq 7/8\), it is positive if \(b\notin\left[k+\left(1-k\right)\left(1-\sqrt{8k-7}\right)/8,k+\left(1-k\right)\left(1+\sqrt{8k-7}\right)/8\right]\).

\(\blacksquare\)