We will prove $\frac{n}{2}$ is a solution for all $n\geq 2$ (the case $n=1$ is trivial). Let $F(x)$ denote $\left( \frac{x^{2n}+1}{2} \right)^\frac{1}{n}$. Calculating the first three derivatives of $F$, we get the following identities: \begin{align*} F(1)&=1,\\ F'(1)&=1,\\ F''(1)&=n,\\ F'''(x)&=-\frac{(2n-1)(2n-2)\cdot 2^{1-\frac{1}{n}}\cdot x^{2n-3} \cdot (x^{2n}-1)}{(x^{2n}+1)^{3-\frac{1}{n}}}. \end{align*}By Taylor's theorem, for every $x\in \mathbb R$ we have $$F(x)-\frac{n}{2}(x-1)^2-x=F'''(w_x)\frac{(x-1)^3}{3!}$$for some $w_x$ between $1$ and $x$. If $x\in [0,1]$, then $F'''(w_x)\geq 0$ and $F(x)-\frac{n}{2}(x-1)^2-x \leq 0$. If $x \in [1, +\infty)$, then $F'''(w_x)\leq 0$ and $F(x)-\frac{n}{2}(x-1)^2-x \leq 0$. If $x<0$, note that $$\frac{n}{2}(x-1)^2+x\geq \frac{n}{2}(-x-1)^2-x \geq F(-x)=F(x),$$where the first inequality holds since it's equivalent to $2x(1-n)\geq 0$, and the second inequality was just proven. Now consider a sequence $a_k=1+\frac{1}{k}$ and let $b_k=w_{a_k}$. For any $c<\frac{n}{2}$, $$F(a_k)-\frac{c}{k^2}-a_k=F'''(b_k)\frac{1}{6k^3}+\frac{\left(\frac{n}{2}-c\right)}{k^2}.$$Multiplying both sides by $k^2$ and taking a limit when $k \rightarrow \infty$, we have $$\lim_{k \to \infty} k^2\left(F(a_k)-\frac{c}{k^2}-a_k\right)=\frac{n}{2}-c>0,$$hence there has to exist a value of $x$ such that $F(x)>c(x-1)^2+x$, which proves that $\frac{n}{2}$ is optimal.

The US found this problem quite hard during our training camp Here's a solution sketch. First, we take the limit as $x \to 1$ of $$\frac{\sqrt[N]{\frac{x^{2N}+1}{2}} - x}{(x-1)^2}$$ We do this by applying L'Hopital's rule twice (!). Next, we prove that the inequality holds for $b_N = \frac{n}{2}$. Take logs, and then derivatives. We end up with a single variable polynomial inequality to prove, which turns out to be quite easy.

Our TST used the first version instead (which is the intended A1 difficulty), the second is A5 or A6 difficulty (according to SL). Here's my solution for the 1st version, which I think is pretty natural to come up with: The answer is $a_n = 2^{n - 1}$ for all $n \in \mathbb{N}$. Note that we have equality for $x = 1$ -- and for $x \not= 1$, we have \[ a_n \ge \frac{\sqrt[2^n]{\frac{x^{2^{n + 1}} + 1}{2}} - x}{(x - 1)^2} \]Let the expression in the reight be $X_n$, and define $A_i = \sqrt[2^i]{\frac{x^{2^{i + 1}} + 1}{2}} + x^{2^{n - i}}$ for all $1 \le i \le n$. Then, we have \begin{align*} X_n &= \frac{\sqrt[2^n]{\frac{x^{2^{n + 1}} + 1}{2}} - x}{(x - 1)^2} \\ &= \frac{\sqrt[2^{n - 1}]{\frac{x^{2^{n + 1}} + 1}{2}} - x^2}{(x - 1)^2 \cdot A_n} \\ &= \frac{\sqrt[2^{n - 2} ]{\frac{x^{2^{n + 1}} + 1}{2}} - x^4}{(x - 1)^2 \cdot A_n \cdot A_{n - 1}} \\ &\vdots \\ &= \frac{\frac{1}{2} (x^{2^n} - 1)^2}{(x - 1)^2 \cdot A_n \cdot A_{n - 1} \cdots A_1} \\ &= \frac{1}{2} \cdot \prod_{i = 1}^{n } \frac{(x^{2^{n - i}} + 1)^2}{A_i} \end{align*}Claim. For any $x \in \mathbb{R}$ and $1 \le i \le n$, we have $A_i \ge \frac{1}{2} (x^{2^{n - i}} + 1)^2$. Proof. By Power Mean inequality, since $1 \le i \le n$, we have \begin{align*} A_i = x^{2^{n - i}} + \sqrt[2^i]{\frac{x^{2^{n + 1}} + 1}{2}} &= x^{2^{n - i}} + \sqrt[2^i]{ \frac{ (x^{2^{n + 1 - i}} )^{2^i} + 1^{2^{i}}}{2}} \\ &\ge x^{2^{n - i}} + \frac{x^{2^{n + 1 - i}} + 1}{2} \\ &= \frac{1}{2} (x^{2^{n - i}} + 1)^2 \end{align*} Using our claim, \begin{align*} X_n &= \frac{1}{2} \cdot \prod_{i = 1}^{n} \frac{(x^{2^{n - i}} + 1)^2}{A_i} \\ &\le \frac{1}{2} \cdot 2^n \\ &= 2^{n - 1} \end{align*}

spartacle wrote: The US found this problem quite hard during our training camp The smallest $C$ is $\tfrac{n}{2}$. Suppose the inequality holds. For $x=-1$, we get $C\ge 1/2$; henceforth assume $x\not = -1$. The main idea is to define $d$ such that $x=\tfrac{1+d}{1-d}$. Then the inequality becomes \[ (1+d)^{2n} + (1-d)^{2n} \le 2\big( (4C-1)d^2+1\big)^n \qquad (\clubsuit) \]upon simplification. This must hold for all $d$, so upon making $d \ll 1$, the $d^2$ coefficient will be the dominating term on both sides. Therefore, the $d^2$ coefficient on the LHS is at most that on the RHS. So \[ 2\binom{2n}{2} \le 2\binom{n}{1}(4C-1) \implies C \ge \frac{n}{2}. \]Now we show that for any fixed $C\ge \tfrac{n}{2}$, the inequality holds. It suffices to show that in $(\clubsuit)$, for each $i=0,1,\ldots,n$, the coefficient of $d^{2i}$ on the LHS is less than or equal to the corresponding coefficient on the RHS. Upon expansion, we find that we want to show \[ 2\binom{2n}{2i} \le 2\binom{n}{i}(4C-1)^i.\]Since $C \ge \tfrac{n}{2}$, it suffices to check \begin{align*} \qquad \frac{(2n)(2n-1)\cdots (2n-2i+1)}{(2i)!} &\le \frac{(n)(n-1)\cdots (n-i+1)}{i!}(2n-1)^i \\ \iff \frac{(2n-1)(2n-3)\cdots (2n-2i+1)}{(2i-1)(2i-3)\cdots (3)(1)} &\le (2n-1)^i. \end{align*}But this is extremely weak and trivially true (in fact, even without the large denominator on the LHS, it is true).

Consider as polynomial

\[ \text{We claim that the answer is} \ \boxed{a_n = 2^{n-1}} \]A strange problem, where $a_n$ itself isn't easily guessed; and the construction is not easily inferred. Saying that, this is a one-idea-problem at heart (see end of $\color{red} \text{Section 2}$ of the Motivation for that). $\color{red} \rule{4.5cm}{2pt}$ $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{\text{What the}} \ 2^n \ \textbf{\text{is for.}}}$ $\color{red} \clubsuit$ $\color{red} \rule{4.5cm}{2pt}$ We will prove this assertion: \[ \sqrt[N]{\dfrac{x^{2N}+1}{2}} \leq 2^{n-1} (x-1)^2+x \]by inducting on $n$. This will prove that $a_n \leq 2^{n-1}$ is sufficient.

$\color{red} \rule{25cm}{0.3pt}$ $\color{red} \spadesuit$ $\color{red} \boxed{\textbf{\text{Proof.}}}$ $\color{red} \spadesuit$ Let we have established the inequality for $n = k$: \[ \sqrt[2^k]{\dfrac{x^{2^{k+1}}+1}{2}} \leq 2^{k-1} (x-1)^2+x \quad \text{or} \quad \dfrac{x^{2^{k+1}}+1}{2} \leq ({2^{k-1} (x-1)^2+x})^{2^{k}} \ \textcolor{red}{\text{(known)}} \]$\forall x \in \mathbb{R}$. Since our objective is to prove \[ \dfrac{x^{2^{k+2}}+1}{2} \leq (2^k (x-1)^2+1)^{2^{k+1}} \ \textcolor{red}{\text{(would like to prove)}} \]seeing that we can make $x^{2^{k+1}}$ into the term that we want $x^{2^{k+2}}$ by substituting $x \rightarrow x^2$; we will do exactly that: \[ \dfrac{x^{2^{k+2}}+1}{2} \leq ({2^{k-1} (x^2-1)^2+x^2})^{2^{k}} \ \textcolor{red}{\text{(known, substituted)}} \]So, if we have proven the following: \[ ({2^{k-1} (x^2-1)^2+x^2})^{2^{k}} \leq (2^k (x-1)^2+1)^{2^{k+1}} \]we are set for life. $\color{cyan} \rule{3.2cm}{2pt}$ $\color{cyan} \clubsuit$ $\boxed{\textbf{Expanding.}}$ $\color{cyan} \clubsuit$ $\color{cyan} \rule{3.2cm}{2pt}$ We will prove \[ (c(x^2-1)^2+x^2) \leq (2c(x-1)^2+x)^2 \]holds for every $c$ so that $4c^2-c \geq 0$. $\color{cyan} \rule{25cm}{0.3pt}$ $\color{cyan} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{cyan} \spadesuit$ Expanding yields that the equation is equivalent to \[ (4c^2-c)x^4-(16c^2-4c)x^3+(8c^2+16c^2-8c+1+2c-1)x^2-(16c^2-4c)x+(4c^2-c) \geq 0 \]or $(4c^2-c)(x-1)^4 \geq 0$. Substituting $c = 2^{k-1}$ and raising it to the power of $2^k$ implies the desired assertion. $\blacksquare$ $\color{red} \rule{9.85cm}{2pt}$ $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{\text{Where we don't need the}} \ 2^n: \textbf{\text{just approximation.}}}$ $\color{red} \clubsuit$ $\color{red} \rule{9.85cm}{2pt}$ Set $m = N$. We will prove that for every $a_m < \dfrac{m}{2}$, there exists $x \in \mathbb{R}$ so that \[ \sqrt[m]{\dfrac{x^{2m}+1}{2}} > a_m (x-1)^2+x \] $\color{red} \rule{25cm}{0.3pt}$ $\color{red} \spadesuit$ $\color{red} \boxed{\textbf{\text{Proof.}}}$ $\color{red} \spadesuit$ We pick $x = 1+\epsilon$ where $\epsilon$ is a very small positive number. The above equation is then equivalent to \[ \dfrac{(1+\epsilon)^{2m}+1}{2} > (a_m \epsilon^2+\epsilon+1)^m \]or \[ \left(1+m\epsilon+\dfrac{\binom{2m}{2}}{2} \epsilon^2+\ldots\right) > \left(1+m\epsilon+\left(\binom{m}{2}+ma_m \right) \epsilon^2+\ldots \right)\]Note that $a_m < \dfrac{m}{2}$ implies $\binom{m}{2}+ma_m < \dfrac{\binom{2m}{2}}{2}$, and the other way around, too. $\color{cyan} \rule{6.35cm}{2pt}$ $\color{cyan} \diamondsuit$ $\boxed{\textbf{Finishing: Thinking in Bases.}}$ $\color{cyan} \diamondsuit$ $\color{cyan} \rule{6.35cm}{2pt}$ Multiplying that expression with $\left(\dfrac{1}{\epsilon}\right)^{2m} = L^{2m}$ where $L$ is large implies the statement \[ \dfrac{\binom{2m}{2}}{2} L^{2m-2}+O(L^{2m-3}) > \left(\binom{m}{2}+ma_m \right) L^{2m-2}+O(L^{2m-3}) \]Since we already know that by the final sentence of $\color{red} \spadesuit$ $\color{red} \boxed{\textbf{\text{Proof}}}$ $\color{red} \spadesuit$ that the $``\text{polynomial}"$ in $L$ on the LHS has a larger leading term than the $``\text{polynomial}"$ in $L$ on the RHS, setting $L \rightarrow \infty$ and $\epsilon = \dfrac{1}{L}$ yields the desired assertion. We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$

I hope this is isomorphic to spartacle's sketch. I do not want to do this again. Until we finish bounding $b_N$, we assume $x>0$. Write the desired inequality as \[\frac{\frac{x^{2N}+1}{2}-\frac{x^N+x^N}{2}}{\sum_{i=0}^{N-1} \sqrt[N]{\frac{x^{2N}+1}{2}}^i\sqrt[N]{\frac{x^N+x^N}{2}}^{N-1-i}}=\sqrt[N]{\frac{x^{2N}+1}{2}}-\sqrt[N]{\frac{x^N+x^N}{2}}\le b_N(x-1)^2.\]Equivalently, it may be written as \[\frac{(x^N-1)^2}{\sum_{i=0}^{N-1} \sqrt[N]{\frac{x^{2N}+1}{2}}^i\sqrt[N]{\frac{x^N+x^N}{2}}^{N-1-i}}=\frac{x^{2N}+1-2x^N}{\sum_{i=0}^{N-1} \sqrt[N]{\frac{x^{2N}+1}{2}}^i\sqrt[N]{\frac{x^N+x^N}{2}}^{N-1-i}}\le 2b_N(x-1)^2.\]Dividing through by $(x-1)^2$, we see that \[\frac{\left(\sum_{i=0}^{N-1}x^i\right)^2}{\sum_{i=0}^{N-1} \sqrt[N]{\frac{x^{2N}+1}{2}}^i\sqrt[N]{\frac{x^N+x^N}{2}}^{N-1-i}}\le 2b_N.\]Equivalently, the inequality may be written as \[\left(\sum_{i=0}^{N-1}x^i\right)^2\le 2b_N\sum_{i=0}^{N-1} \sqrt[N]{\frac{x^{2N}+1}{2}}^i\sqrt[N]{\frac{x^N+x^N}{2}}^{N-1-i}.\]We claim the correct minimum value of $b_N$ is $\frac N2$: remark that the inequality above is equivalent to the original inequality. It is clear that $b_N\ge \frac N2$ by setting $x=1$. Now, we raise both sides to the $N$-th power in the original equation to see we wish to show \[x^{2N}+1\le 2\left(\frac N2(x-1)^2+x\right)^N = 2\left(\frac N2x^2+\frac N2 - (N-1)x\right)^N = \frac{1}{2^{N-1}}\left(Nx^2-2(N-1)x+N\right)^N.\]Note the result is clear for $x\le 0$ by replacing it with $x\ge 0$, so WLOG $x\ge 0$. Moreover, the result is certainly true for $x=0$: note $1\le \frac{N^N}{2^{N-1}} = N\cdot (\frac N2)^{N-1}$. Now take the rewrite and write it as \[\sum_{i=1}^N ix^{i-N}+ix^{N-i}\le N\sum_{i=0}^{N-1}\sqrt[N]{\frac{x^N+\frac{1}{x^N}}{2}}^i.\]Remark that we are automatically done if $N=1$, so assume $N\ge 2$. We claim that \[\left(\frac{2^t+2^{-t}}{2}\right)^{\frac 1t}\]is an increasing function in $t$. This would imply that $\sqrt[N]{\frac{x^N+x^{-N}}{2}} \ge \sqrt[N-1]{\frac{x^{N-1}+x^{1-N}}{2}}$ which suffices because then we would have \[N\sum_{i=0}^{N-1}\sqrt[N]{\frac{x^N+\frac{1}{x^N}}{2}}^i\ge N\sum_{i=0}^{N-1} \frac{x^i + x^{-i}}{2}\]and then the fact that $x^i+x^{-i}\ge x^{i-1}+x^{-i+1}$ for $i>0$ finishes. To be precise, if $i\le N-1$ we get $\frac N2\cdot i \ge 1+2+\cdots+i=i(i+1)/2$ (analyzing the number of terms $x^n$ with $|n|\ge N-i$) and if $i=N$ the result carries over because the sides are equal at $x=1$. Indeed, this is an increasing function because it has derivative of its logarithm equal to \[\left(\frac 1t \log\left(\frac{2^t+2^{-t}}{2}\right)\right)' = -\frac 1{t^2}\log\left(\frac{2^t+2^{-t}}{2}\right) + \frac 1t \cdot \frac{\log 2\cdot 2^t-\log 2\cdot 2^{-t}}{2^t+2^{-t}}.\]So long as $t>0$, which we can safely assume, this is positive if and only if \[t\cdot \log 2\cdot \frac{2^t-2^{-t}}{2^t+2^{-t}}>\log\left(\frac{2^t+2^{-t}}{2}\right) = \log(2^t+2^{-t}) - \log 2 = t\log 2 - \log 2 + \log(1+2^{-2t}).\]That is, we are done as long as \[t\log 2 - 2t\log 2 \cdot \frac{2^{-2t}}{1+2^{-2t}} > t\log 2 - \log 2 + \log(1+2^{-2t}).\]So we must argue that \[2t\log 2 \cdot \frac{2^{-2t}}{1+2^{-2t}} < \log 2 - \log(1+2^{-2t}).\]Let $2^{2t} = x$. Then we wish to show \[\log x\cdot \frac{1}{1+x} < \log 2 - \log (1+1/x).\]This is equivalent to proving \[\log 2 > \log(1+1/x)+\log x\cdot \frac{1}{1+x}.\]Since equality would hold at $x=1$ and we have $x>1$, it is enough to show the derivative of the expression on the right is negative. It has derivative \[-\frac 1{x^2}\cdot\frac{1}{1+\frac 1x} + \frac 1x\cdot \frac{1}{1+x} - \log x\cdot \frac{1}{(1+x)^2}.\]Indeed, this derivative is always negative, so we are done.

Bit of a dumb problem. Here's a solution for version 1. The answer is $\tfrac{N}{2}=2^{n-1}$. Note that for $x \neq 1$, we require $$a_n\geq \frac{\sqrt[N]{\frac{x^{2N}+1}{2}}-x}{(x-1)^2}.$$Using L'Hopital's twice, we have $$\lim_{x \to 1} \frac{\sqrt[N]{\frac{x^{2N}+1}{2}}-x}{(x-1)^2}=\lim_{x \to 1} \frac{\left(\frac{x^{2N}+1}{2}\right)^{1/N-1}x^{2N-1}-1}{2x-2}=\lim_{x \to 1} \frac{(2N-1)x^{2N-2}\left(\frac{x^{2N}+1}{2}\right)^{1/N-2}+(N-1)x^{4n-2}\left(\frac{x^{2N}+1}{2}\right)^{1/N-1}}{2}=\frac{N}{2}$$or something like that, which proves that the answer is a lower bound. To prove that it works, se use induction, with the base case of $n=0 \implies N=1$ being clear. Now suppose that $N$ works—I will prove that $2N$ works as well. Indeed, we have $$\sqrt[2N]{\frac{x^{4N}+1}{2}}-x=\sqrt[2N]{\frac{x^{4N}+1}{2}}+\sqrt[2N]{x^{2N}}-2x \leq 2\sqrt[2N]{\frac{x^{4N}+x^{2N}+1}{4}}-2x=2\left(\sqrt[N]{\frac{x^{2N}+1}{2}}-x\right)$$by Jensen's inequality, as $\sqrt[2N]{x}$ is concave. This implies $$\sqrt[2N]{\frac{x^{4N}+1}{2}}-x\leq 2\cdot \frac{N}{2}(x-1)^2=N(x-1)^2,$$which is the inequality for $2N$. $\blacksquare$

Let $y=1$. Then, the inequality becomes $\sqrt[N]{\frac{x^{2N}+y^{2N}}2}\leq b_N(x-y)^2+xy$. This is homogenous with degree $2$, so if $x\neq y$, we can assume without loss of generality that $x=k-1$ and $y=k+1$ by scaling. If $x=y$, then the inequality is obviously true. Now, making the substitution, we get $$\sqrt[N]{\frac{(k-1)^{2N}+(k+1)^{2N}}2}\leq4b_N+k^2-1,$$or $(k-1)^{2N}+(k+1)^{2N}\leq2(4b_N+k^2-1)^N$. Expanding, we get $$2\sum_{i=0}^N\binom Nik^{2i}(4b_N-1)^{N-i}\geq2\sum_{i=0}^N\binom{2N}{2i}k^{2i},$$so $$\sum_{i=0}^Nk^{2i}\left(\binom Ni(4b_N-1)^{N-i}-\binom{2N}{2i}\right)\geq0.$$We have that the $k^{2N}$ coefficient of this is zero and the $k^{2N-2}$ coefficient is $N(4b_N-1)-N(2N-1)$. If the inequality is true, then $4b_N-2N\geq0$, so $b_N\geq\frac N2$. Now, we will show that $b_N=\frac N2$ works. It suffices to show $$\sum_{i=0}^Nk^{2i}\left(\binom Ni(2N-1)^{N-i}-\binom{2N}{2i}\right)\geq0.$$We claim that $\binom Ni(2N-1)^{N-i}\geq\binom{2N}{2i}$ for all $i\leq N$, which proves the inequality as $k^{2i}$ is always nonnegative. We have \begin{align*} \frac{\frac{N!}{i!(N-i)!}(2N-1)^{N-i}}{\frac{(2N)!}{(2i)!(2N-2i)!}}&=\frac{N!(2N-1)^{N-i}(2i)!(2N-2i)!}{(2N)!i!(N-i)!}\\ &\geq\frac{N!(2N-1)^{N-i}(2i)!(N-i)!2^{N-i}}{(2N)!i!(N-i)!}\\ &=\frac{N!(2N-1)^{N-i}(2i)!2^{N-i}}{(2N)!i!}\\ &=2^{N-i}(2N-1)^{N-i}\frac{2^i(1\cdot3\cdot\ldots\cdot(2i-1))}{2^N(1\cdot3\cdot\ldots\cdot(2N-1))}\\ &=\frac{(2N-1)^{N-i}}{(2i+1)\cdot(2i+3)\cdot\ldots\cdot(2N-1)}\\ &\geq1. \end{align*}Therefore, each coefficient is nonnegative, so the entire sum is nonnegative. This implies that the smallest $b_N$ satisfying the inequality is $\boxed{\frac N2}$.

i notice that the substitution x=k-1 and y=k+1 or the x=(k+1)/(k-1) substitution kills quickly; what was the motivation for this? my progress was noting that x=k+1 sub might do something, so i am mostly asking where motivation for x=(k+1)/(k-1) comes from

Let $x=\tfrac{a+b}{a-b}$ for some $a,b$. Then, the inequality becomes \begin{align*}\ &\sqrt[N]{\frac{(a+b)^{2N}+(a-b)^{2N}}{2(a-b)^N}} \le a_{n}\left(\frac{2b}{a-b}\right)^{2}+\frac{a+b}{a-b} \\ \iff &\frac{(a+b)^{2N}+(a-b)^{2N}}{2(a-b)^N} \le \left(\frac{4a_nb^2+a^2-b^2}{(a-b)^2}\right)^N \\ \iff &(a+b)^{2N}+(a-b)^{2N} \le ((4a_n-1)b^2+a^2)^N. \end{align*}Since that is homogeneous, set $b=1$ to get $(a+1)^{2N}+(a-1)^{2N}\le (a^2+4a_n-1)^N.$ Now, we can see this as the following comparison of polynomial: \[\sum_{i=0}^{N}{\binom{2N}{2i}a^{2i}}\le \sum_{i=0}^{N}\binom {N}{i}a^{2i}(4a_n-1)^{N-i}\]Note that if we want this to be true for all $a$, the largest differing coefficients must have the LHS smaller. The largest coefficient, $a^{2i}$, is the same. The next largest, $a^{2i-2}$, we have \[\frac{2N(2N-1)}{2}\le N\cdot (4a_n-1)\]so we must have $a_n=2^{n-1}=\tfrac{N}{2}$ and we know that $\tfrac{N}{2}$ works by comparing the rest of the terms of the polynomial.

Giving a solution for the first version of the problem. Nice inequality We claim the answer is $2^{n-1}$. To get the lower bound, we can take the limit $x$ tends to $1$. To prove that this $a_n$ works, we induct on $n$. The base case $a = 0$ is trivial (it is an equality in this case). Now, suppose the statement is true for $n-1$. Then, \[ \sqrt[N]{\frac{x^{2 N}+1}{2}} + \sqrt[N]{x^N} \leqslant 2\sqrt[N]{\frac{x^{2N}+x^N+1}{2}} \] But the thing under the radical on the right factorises to $(x^N+1)^2$, whose ${N}$th root by the inductive hypothesis is atmost $a_{n-1}(x-1)^2+x$. Thus we get $a_{n} \leq 2a_{n-1}$ and we are done.

Let me try a motivated solution using roots of unity and AM-GM. Let $\omega=\textrm{cis}\left(\frac{\pi}{2N}\right)$ be a primitive $4N$-th root of unity. Then we have \begin{align*} x^{2N}+1&=\prod_{i=1}^{2N}(x-\omega^{2i-1})\\ &=\prod_{i=1}^{N}(x-\omega^{2i-1})(x-\omega^{1-2i})\\ &=\prod_{i=1}^{N}\left(x^2-2\cos\left(\frac{(2i-1)\pi}{2N}\right)x+1\right). \end{align*}This temps us to use AM-GM. Equality holds when $x=1$. Then we have \begin{align*} x^2-2\cos\left(\frac{(2i-1)\pi}{2N}\right)x+1&=2-2\cos\left(\frac{(2i-1)\pi}{2N}\right)\\ &=2\left[\cos 0-\cos\left(\frac{(2i-1)\pi}{2N}\right)\right]\\ &=4\sin\left(\frac{(2i-1)\pi}{4N}\right)^2. \end{align*}This also proves the identity $$x^{2N}+1=\prod_{i=1}^{N}\left[(x-1)^2+4\sin\left(\frac{(2i-1)\pi}{4N}\right)^2x\right].\qquad(*)$$Now let $t_i=4\sin\left(\frac{(2i-1)\pi}{4N}\right)^2>0$. Substituting $x=1$ in $(*)$ implies that $\prod_{i=1}^N t_i=2$. Differentiating $(*)$ twice and substituting $x=1$ implies that $$2N(2N-1)=2\sum_{j=1}^N\prod_{i=1,i\ne j}^Nt_i+N(N-1)\prod_{i=1}^N t_i.$$Hence $\sum_{j=1}^N\prod_{i=1,i\ne j}^Nt_i=N^2$ so $\sum_{i=1}^N\frac{1}{t_i}=\frac{N^2}{2}$. Now, we can reformulate $(*)$ as $$\frac{x^{2N}+1}{2}=\frac{x^{2N}+1}{\prod_{i=1}^N t_i}=\prod_{i=1}^{N}\left[\frac{(x-1)^2}{t_i}+x\right].$$Now, by AM-GM, $$\frac{x^{2N}+1}{2}\le\left[\frac{(x-1)^2}{N}\sum_{i=1}^N\frac{1}{t_i}+x\right]^N=\left[\frac{N}{2}(x-1)^2+x\right]^N.$$This proves that $a_N=\frac{N}{2}$ works. To prove that $b_N\ge\frac{N}{2}$, we see that: \begin{align*} b_N(x-1)^2&\ge\sqrt[N]{\frac{x^{2N}+1}{2}}-x\\ &=\frac{\frac{x^{2N}+1}{2}-x^N}{\sum_{i=0}^{N-1}\left(\frac{x^{2N}+1}{2}\right)^{\frac{i}{N}}x^{N-1-i}}\\ &=\frac{(x^N-1)^2}{2\sum_{i=0}^{N-1}\left(\frac{x^{2N}+1}{2}\right)^{\frac{i}{N}}x^{N-1-i}}\\ &=\frac{(x-1)^2\left(\sum_{i=1}^{N-1}x^i\right)^2}{2\sum_{i=0}^{N-1}\left(\frac{x^{2N}+1}{2}\right)^{\frac{i}{N}}x^{N-1-i}}. \end{align*}Now taking $x$ close to $1$ implies that $b_N\ge\frac{N}{2}$.

average famat kiddo... We claim the answer is $\frac N2$ (for version 2). Preliminaries: We consider the function $f$ that is the difference of the right and left hand sides. The derivative of this function is $2b_N(x - 1) + 1 - 2^{1 - \frac 1N} x^{2N - 1} (x^{2N} + 1)^{-1 + \frac 1N}$. The second derivative of this function is $2b_N - 2^{1 - \frac 1N} (-2(N - 1)x^{2N - 1} x^{2N - 1} (x^{2N} + 1)^{-2 + \frac 1N} + (2N - 1)x^{2N -2} (x^{2N} + 1)^{-1 + \frac 1N}) = 2b_N - 2^{1 - \frac 1N}x^{2N - 2}(2Nx^{2N} ) = 2b_N - 2^{1 - \frac 1N} x^{2N - 2} (x^{2N} + 1)^{-2 + \frac 1N} (-2(N - 1) x^{2N} + (2N - 1)(x^{2N} + 1)) = 2b_N - 2^{1 - \frac 1N} x^{2N - 2}(x^{2N} + 1)^{-2 + \frac 1N} (-2Nx^{2N} + 2x^{2N} + 2Nx^{2N} + 2N - x^{2N} - 1) = 2b_N - 2^{1 - \frac 1N} x^{2N - 2}(x^{2N} + 1)^{-2 + \frac 1N}(x^{2N} + 2N - 1) $. Proof of minimality: Observe both $f$ and $f'$ are $0$ at $x = 1$, so to prove $x = 1$ is a relative minimum, we need to the second derivative to be nonnegative at $x = 1$, which gives $2b_N \ge 2^{1 - \frac 1N} 2^{-2 + \frac 1N} (2N) = N$, so $b_N \ge \frac N2$ as desired. Proof of sufficiency: This is completely obvious for $N = 1$, so assume $N \ge 2$. Observe $f(0) > 0$, so if we show the derivative of the function is negative for all $x < 0$, the inequality will hold for all $x \le 0$. Of course, the derivative of the function can be bounded as $Nx - N + 1 - 2^{1 - \frac 1N} x^{2N - 1}(x^{2N} + 1)^{-1 + \frac 1N}$ for negative $x$ is at most $Nx - N + 1 - 2^{1 - \frac 1N}x$ which is at most $Nx - N + 1 - 2x$, which is always negative. Now for $0 < x < 1$, we show that the third derivative of the function is negative, meaning that the second derivative was decreasing over that interval, since the second derivative is zero at $1$ this forces that the second derivative was always positive over that interval, which means that the derivative was always increasing over that interval, since the derivative was $0$ at $1$ this implies that the derivative was always negative over that interval, which means that $f$ was always decreasing over that interval, meaning that $f(x) > f(1)$ for $0 < x < 1$. Taking the third derivative gives $-2^{1 - \frac 1N} ((-2 + \frac 1N) (x^{2N} + 1)^{-3 + \frac 1N} 2Nx^{2N - 1} x^{2N - 2}(x^{2N} + 2N - 1) +(x^{2N} + 1)^{-2 + \frac 1N}((4N -2) x^{4N - 3} + (2N - 1)(2N - 2) x^{2N - 3}))$, which is $-2^{1 - \frac 1N} x^{2N - 3} (x^{2N + 1} +1)^{-3 + \frac 1N}(2(1 - 2N) x^{2N}(x^{2N} + 2N - 1)) + (x^{2N} + 1)((4N - 2)x^{2N }) + (2N - 1)(2N - 2) )$. This has opposite sign to $2(1 - 2N)x^{4N} - x^{2N}2(2N -1)^2 + (4N - 2)x^{4N} + (4N^2 - 6N + 2 + 4N - 2)x^{2N} + (2N - 1)(2N - 2)$ which is equivalent to $x^{2N} (-2(2N - 1)^2 + (4N^2 - 2N)) + (2N - 1)(2N - 2) = x^{2N} (-4N^2 + 6N - 2) + (4N^2 - 6N + 2) = (1 - x^{2N})(2N - 1)(2N - 2)$. Thus for $0 \le x \le 1$, we can easily see that this expression has positive sign, meaning that the third derivative has negative sign. We also claim that for showing the inequality holds for $x > 1$, it is sufficient to show the third derivative is positive for $x > 1$: this is because the third derivative being positive implies the second derivative is increasing, since the second derivative is zero at $x = 1$ it implies that the second derivative is positive for all $x > 1$, which implies the first derivative is increasing for all $x > 1$, since the first derivative is zero at $x = 1$ it implies the first derivative is positive for all $x > 1$, which implies that the function is increasing from $x > 1$, since $f(1) = 0$ this is sufficient. This is also obvious upon considering $(1 - x^{2N}) (2N - 1)(2N - 2)$ is negative, implying that the third derivative is positive over this interval.

Claim: We have $b_n \ge \frac{n}{2}$. Proof. Consider the function \[ b_n(x-1)^2 + x - \left( \frac{x^{2n}+1}{2} \right)^{\frac 1n} \]it must always be nonnegative, yet it is zero at $x = 1$. Then its second derivative mut be nonnegative at $x = 1$. The first derivative is \[ 2b_n(x-1) + 1 - x^{2n-1} \left( \frac{x^{2n}+1}{2} \right)^{\frac 1n - 1} \]the second derivative is \[ 2b_n - x^{2n-1} \cdot (1-n) \cdot x^{2n-2} \left( \frac{x^{2n}+1}{2} \right)^{\frac 1n - 2} - (2n-1)x^{2n-2} \cdot \left( \frac{x^{2n}+1}{2} \right)^{\frac 1n - 1} \]Setting $x = 1$, we find that \[ 2b_n - (1-n) - (2n - 1) \ge 0 \implies b_n \ge \frac{n}{2} \]as needed. $\blacksquare$ We now prove that $b_n = \frac{n}{2}$ works. Let $y = 1$, and so the inequality rewrites as \[ \left( \frac{x^{2n} + y^{2n}}{2} \right)^{\frac 1n} \le \frac n2(x-y)^2 + xy \]get rid of the condition $y = 1$. Now the inequality is homogenous, so we may assume via scaling that $x + y = 2$ (check the case $x = -y$ easily works). Set $x = 1 + t$ and $y = 1 - t$, and so \[ \left( \frac{(1+t)^{2n} + (1-t)^{2n}}{2} \right)^{\frac 1n} \le \frac n2 (2t)^2 + (1-t^2) \]we do this because then \[ \frac{(1+t)^{2n} + (1-t)^{2n}}{2} = t^{2n} + \binom{2n}{2n-2}t^{2n-2} + \dots + 1 \]so it suffices to show that \[ t^{2n} + \binom{2n}{2n-2}t^{2n-2} + \dots + 1 \le ( (2n-1)t^2 + 1)^n \]this is equivalent to showing that for all integers $i$ with $1 \le i \le n$, \[ \binom{n}{i} (2n-1)^i \ge \binom{2n}{2i} \]\[ \iff (2n-1)^i \ge \frac{ \frac{(2n)!}{n!} }{ \frac{(2i)!}{i!} \frac{(2n-2i)!}{(n-i)!} } \]\[ \iff (2n-1)^i \ge \frac{(2n-1)(2n-3) \cdots (1)}{(2i-1)(2i-3)\cdots 1 \cdot (2n-2i-1) \cdot (2n-2i-3) \cdots 1} \]\[ \iff (2n-1)^i \ge \frac{(2n-1)(2n-3) \cdots (2n-2i+1)}{(2i-1)(2i-3)\cdots 1} \]which is clearly true. $\blacksquare$