Version 1. Let $n$ be a positive integer, and set $N=2^{n}$. Determine the smallest real number $a_{n}$ such that, for all real $x$,  
$$ \sqrt[N]{\frac{x^{2 N}+1}{2}} \leqslant a_{n}(x-1)^{2}+x $$  
Version 2. For every positive integer $N$, determine the smallest real number $b_{N}$ such that, for all real $x$,  
$$ \sqrt[N]{\frac{x^{2 N}+1}{2}} \leqslant b_{N}(x-1)^{2}+x $$  
(Ireland)
1 (for Version 2). Like in   First of all, $\mathcal{I}(N, 0)$ is obvious. Further, if $x>0$, then the left hand sides of $\mathcal{I}(N,-x)$ and $\mathcal{I}(N, x)$ coincide, while the right hand side of $\mathcal{I}(N,-x)$ is larger than that of $\mathcal{I}(N, x)$ (their difference equals $2(N-1) x \geqslant 0)$. Therefore, $\mathcal{I}(N,-x)$ follows from $\mathcal{I}(N, x)$. So, hereafter we suppose that $x>0$.  Divide $\mathcal{I}(N, x)$ by $x$ and let $t=(x-1)^{2} / x=x-2+1 / x$; then $\mathcal{I}(N, x)$ reads as  $$ f_{N}:=\frac{x^{N}+x^{-N}}{2} \leqslant\left(1+\frac{N}{2} t\right)^{N} $$  The key identity is the expansion of $f_{N}$ as a polynomial in $t$ :