Problem (Algebra):
Let $n$ be a positive integer, and set $N=2^{n}$. Determine the smallest real number $a_{n}$ such that, for all real $x$,

$$
\sqrt[N]{\frac{x^{2 N}+1}{2}} \leqslant a_{n}(x-1)^{2}+x
$$

Solution:
First of all, assume that $a_{n}<N / 2$ satisfies the condition. Take $x=1+t$ for $t>0$, we should have

$$
\frac{(1+t)^{2 N}+1}{2} \leqslant\left(1+t+a_{n} t^{2}\right)^{N}
$$

Expanding the brackets we get

$$
\left(1+t+a_{n} t^{2}\right)^{N}-\frac{(1+t)^{2 N}+1}{2}=\left(N a_{n}-\frac{N^{2}}{2}\right) t^{2}+c_{3} t^{3}+\ldots+c_{2 N} t^{2 N}
\tag{1}
$$

with some coefficients $c_{3}, \ldots, c_{2 N}$. Since $a_{n}<N / 2$, the right hand side of (1) is negative for sufficiently small $t$. A contradiction.

It remains to prove the following inequality

$$
\text{let   }\mathcal{I}(N, x)\text{   be   }\sqrt[N]{\frac{1+x^{2 N}}{2}} \leqslant x+\frac{N}{2}(x-1)^{2}
$$

where $N=2^{n}$.

Use induction in $n$. The base case $n=0$ is trivial: $N=1$ and both sides of $\mathcal{I}(N, x)$ are equal to $\left(1+x^{2}\right) / 2$. For completing the induction we prove $\mathcal{I}(2 N, x)$ assuming that $\mathcal{I}(N, y)$ is established for all real $y$. We have

$$
\begin{aligned}
\left(x+N(x-1)^{2}\right)^{2} & =x^{2}+N^{2}(x-1)^{4}+N(x-1)^{2} \frac{(x+1)^{2}-(x-1)^{2}}{2} \\
& =x^{2}+\frac{N}{2}\left(x^{2}-1\right)^{2}+\left(N^{2}-\frac{N}{2}\right)(x-1)^{4} \geqslant x^{2}+\frac{N}{2}\left(x^{2}-1\right)^{2} \geqslant \sqrt[N]{\frac{1+x^{4 N}}{2}}
\end{aligned}
$$

where the last inequality is $\mathcal{I}\left(N, x^{2}\right)$. Since

$$
x+N(x-1)^{2} \geqslant x+\frac{(x-1)^{2}}{2}=\frac{x^{2}+1}{2} \geqslant 0
$$

taking square root we get $\mathcal{I}(2 N, x)$. The inductive step is complete.

Final answer: $\frac{N}{2}$