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Now we have to distinguish two subcases of Case 1, with further subsubcases of Subcase 1:

Now we have to distinguish two subcases of Case 1, with further subsubcases of Subcase 1.1:

It turns out that \begin{equation} (G'_n)|_{n=\ell_1/2;\,x=1-tv,\,y=1-v;\,t=(1/3)/(1+r),\,v=(1-y_1)/(1+s)}\; (1 + r)^{10} (1 + s)^{19} \end{equation} and \begin{equation} (G'_n)|_{n=\ell_1/2;\,x=1-tv,\,y=1-v;\,t=(1/3)/(1+r),\,v=(1-y_1)/(1+s)}\; (1 + r)^{10} (1 + s)^{19}\; q \end{equation} are each a polynomial in $r,s$ with all coefficients nonnegative, where \begin{multline*} q:= (1 + r)^{10} (1 + s)^{19} \\ \times(11673186598630578538556565100133681446610566511878526881 \\ + 16777216000000000000000000000000000000000000000000000000 s)^2 \\ \times\big(31853088866210192846185521700044560482203522170626175627 \\ + 33554432000000000000000000000000000000000000000000000000 (r + s + r s)\big)^2. \end{multline*}

It turns out that \begin{equation} (G'_n)|_{n=\ell_1/2;\,x=1-tv,\,y=1-v;\,t=(1/3)/(1+r),\,v=(1-y_1)/(1+s)}\; (1 + r)^{10} (1 + s)^{19} \end{equation} and \begin{equation} (G'_n)|_{n=\ell_2/2;\,x=1-tv,\,y=1-v;\,t=(1/3)/(1+r),\,v=(1-y_1)/(1+s)}\; (1 + r)^{10} (1 + s)^{19}\; q \end{equation} are each a polynomial in $r,s$ with all coefficients nonnegative, where \begin{multline*} q:= (1 + r)^{10} (1 + s)^{19} \\ \times(11673186598630578538556565100133681446610566511878526881 \\ + 16777216000000000000000000000000000000000000000000000000 s)^2 \\ \times\big(31853088866210192846185521700044560482203522170626175627 \\ + 33554432000000000000000000000000000000000000000000000000 (r + s + r s)\big)^2. \end{multline*}

In Case 2, $y$ is "close" to $1$ and hence, in view of inequality $y<x^{24}$, so is $x$: $x>y_1^{1/12}=x_1=0.985$. So, it is a bit more convenient here to change variables $x, y$ to "small variables" $u:=1-x$ and $v:=1-y$. Then $0<v<1-y_1\approx0.304$ and $1-u=x>y^{1/24}=(1-v)^{1/24}>1-v/3$, whence $u<v/3$. Let now $t:=u/v$, so that $0<t<1/3$. It follows, in Case 2, that $t=(1/3)/(1+r)$ and $v=(1-y1)/(1+s)$ for some $r\ge0$ and $s\ge0$.

In Case 2, $y$ is "close" to $1$ and hence, in view of inequality $y<x^{24}$, so is $x$: $x>y_1^{1/12}=x_1=0.985$. So, it is a bit more convenient here to change variables $x, y$ to "small variables" $u:=1-x$ and $v:=1-y$. Then $0<v<1-y_1\approx0.304$ and $1-u=x>y^{1/24}=(1-v)^{1/24}>1-v/3$, whence $u<v/3$. Let now $t:=u/v$, so that $0<t<1/3$. It follows, in Case 2, that $t=(1/3)/(1+r)$ and $v=(1-y_1)/(1+s)$ for some $r\ge0$ and $s\ge0$.