Lemma: If $x$ and $y$ are positive reals such that $xy=1$, then $4+x^2+y^2 \ge 3(x+y)$, with equality if and only if $x=y=1$. Proof: Note that $4+x^2+y^2=(x+y)^2+2=((x+y)-2)((x+y)-1)+3(x+y) \ge 3(x+y)$ since $x+y \ge 2\sqrt{xy}=2$. Equality holds if and only if $x+y=2 \iff x=y=1$. Let $x_i = a_{2i-1}$ and $y_i = a_{2i}$ for all $i=1,2,\dots,n$. We see that there are $2^n$ distinct terms in the form of $t_1t_2\dots t_n$, where each $t$ can either be an $x$ or a $y$. For every $T=t_1t_2\dots t_n$, there is a unique $T’=t_1’t_2’\dots t_n’$ such that $\{t_i,t_i’\}=\{x_i,y_i\}$ for all $i=1,2,\dots,n$. So $T \cdot T’=t_1t_1’\dots t_nt_n’=x_1y_1\dots x_ny_n = a_1a_2\dots a_{2n-1}a_{2n}=1$. Thus, we can label those $2^n$ terms $T_1, T_2, \dots, T_{2^n}$ such that $T_{2i-1} \cdot T_{2i} = 1$ for all $i=1,2,\dots, 2^{n-1}$. We see that \begin{align*} 2^{n+1}+(a_1^2+a_2^2)(a_3^2+a_4^2)\dots(a_{2n-1}^2+a_{2n}^2) &= 2^{n+1}+(x_1^2+y_1^2)(x_2^2+y_2^2)\dots(x_n^2+y_n^2) \\ &= 2^{n+1}+T_1^2+T_2^2+\dots+T_{2^n}^2 \\ &= (4+T_1^2+T_2^2)+\dots+(4+T_{2^n-1}^2+T_{2^n}^2) \\ &\ge 3(T_1+T_2)+\dots+3(T_{2^n-1}+T_{2^n}) \\ &= 3(T_1+\dots+T_{2^n}) \\ &= 3(x_1+y_1)\dots(x_n+y_n) \\ &= 3(a_1+a_2)(a_3+a_4)\dots(a_{2n-1}+a_{2n}) \end{align*}Equality holds if and only if $T_{2i-1}=T_{2i}=1$ for all $i=1,2,\dots, 2^{n-1}$ which means $T_1=T_2=\dots=T_{2^n}=1$. This would imply that $x_i=y_i$ for all $i=1,2,\dots, n$ and $a_{2i-1}=a_{2i}$ for all $i=1,2,\dots, n$. It is easy to check equality holds when $a_{2i-1}=a_{2i}$ for all $i=1,2,\dots, n$.