, 2021

The answer is $b_N = N/2$.

Solution 1. First of all, assume that $b_N < N/2$ satisfies the condition. Take $x = 1 + t$ for $t > 0$, we should have
$$\frac{(1 + t)^{2N} + 1}{2} \le (1 + t + b_N t^2)^N.$$
Expanding the brackets we get
$$(1 + t + b_N t^2)^N - \frac{(1 + t)^{2N} + 1}{2} = \left(Nb_N - \frac{N^2}{2}\right)t^2 + c_3 t^3 + \dots + c_{2N} t^{2N} \quad (1)$$
with some coefficients $c_3, \dots, c_{2N}$. Since $b_N < N/2$, the right hand side of (1) is negative for sufficiently small $t$. A contradiction.

Now we denote $\mathcal{I}(N, x)$ for the inequality. It remains to prove the inequality $\mathcal{I}(N, x)$ with $b_N = N/2$ for an arbitrary positive integer $N$.

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 $2(N-1)x \ge 0$). 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
$$f_N := \frac{x^N + x^{-N}}{2} \le \left(1 + \frac{N}{2}t\right)^N. \qquad (2)$$
The key identity is the expansion of $f_N$ as a polynomial in $t$:
Lemma.
$$f_N = N \sum_{k=0}^{N} \frac{1}{N+k} \binom{N+k}{2k} t^k. \qquad (3)$$
---
Proof. Apply induction on $N$. We will make use of the straightforward recurrence relation
$$f_{N+1} + f_{N-1} = (x + 1/x)f_N = (2 + t)f_N. \quad (4)$$
The base cases $N = 1, 2$ are straightforward:
$$f_1 = 1 + \frac{t}{2}, \quad f_2 = \frac{1}{2}t^2 + 2t + 1.$$
For the induction step from $N-1$ and $N$ to $N+1$, we compute the coefficient of $t_k$ in $f_{N+1}$ using the formula $f_{N+1} = (2+t)f_N - f_{N-1}$. For $k=0$ that coefficient equals 1, for $k > 0$ it equals
$$\begin{align*} & 2 \frac{N}{N+k} \binom{N+k}{2k} + \frac{N}{N+k-1} \binom{N+k-1}{2k-2} - \frac{N-1}{N+k-1} \binom{N+k-1}{2k} \\ &= \frac{(N+k-1)!}{(2k)!(N-k)!} \left( 2N + \frac{2k(2k-1)N}{(N+k-1)(N-k+1)} - \frac{(N-1)(N-k)}{N+k-1} \right) \\ &= \frac{(N+k-1)!}{(2k)!(N-k+1)!} \left( 2N(N-k+1) + 3kN + k - N^2 - N \right) = \frac{\binom{N+k+1}{2k}}{(N+k+1)} (N+1), \end{align*}$$
that completes the induction. $\square$
Turning back to the problem, in order to prove (2) we write
$$(1 + \frac{N}{2}t)^N - f_N = \left(1 + \frac{N}{2}t\right)^N - N \sum_{k=0}^{N} \frac{1}{N+k} \binom{N+k}{2k} t^k = \sum_{k=0}^{N} \alpha_k t^k,$$
where
$$\begin{align*} \alpha_k &= \left(\frac{N}{2}\right)^k \binom{N}{k} - \frac{N}{N+k} \binom{N+k}{2k} \\ &= \left(\frac{N}{2}\right)^k \binom{N}{k} \left(1 - 2^k \frac{(1+1/N)(1+2/N)\cdots(1+(k-1)/N)}{(k+1)\cdots(2k)}\right) \\ &\ge \left(\frac{N}{2}\right)^k \binom{N}{k} \left(1 - 2^k \frac{2 \cdot 3 \cdots k}{(k+1)\cdots(2k)}\right) = \left(\frac{N}{2}\right)^k \binom{N}{k} \left(1 - \prod_{j=1}^k \frac{2j}{k+j}\right) \ge 0, \end{align*}$$
and (2) follows.

Solution 2. Here we present another proof of the inequality (2) for $x > 0$, or, equivalently, for $t = (x - 1)^2/x \ge 0$. Instead of finding the coefficients of the polynomial $f_N = f_N(t)$ we may find its roots, which is in a sense more straightforward. Note that the recurrence (4) and the initial conditions $f_0 = 1, f_1 = 1 + t/2$ imply that $f_N$ is a polynomial in $t$ of degree $N$. It also follows by induction that $f_N(0) = 1, f'_N(0) = N^2/2$: the recurrence relations read as $f_{N+1}(0) + f_{N-1}(0) = 2f_N(0)$ and $f'_{N+1}(0) + f'_{N-1}(0) = 2f'_N(0) + f_N(0)$, respectively.
Next, if $x_k = \exp\left(\frac{i\pi(2k-1)}{2N}\right)$ for $k \in 1, 2, \dots, N$, then
$$-t_k := 2 - x_k - \frac{1}{x_k} = 2 - 2 \cos \frac{\pi(2k-1)}{2N} = 4 \sin^2 \frac{\pi(2k-1)}{4N} > 0$$
and
$$f_N(t_k) = \frac{x_k^N + x_k^{-N}}{2} = \frac{\exp\left(\frac{i\pi(2k-1)}{2}\right) + \exp\left(-\frac{i\pi(2k-1)}{2}\right)}{2} = 0.$$
So the roots of $f_N$ are $t_1, \dots, t_N$ and by the AM-GM inequality we have
$$\begin{aligned} f_N(t) &= \left(1 - \frac{t}{t_1}\right)\left(1 - \frac{t}{t_2}\right)\dots\left(1 - \frac{t}{t_N}\right) \le \left(1 - \frac{t}{N}\left(\frac{1}{t_1} + \dots + \frac{1}{t_N}\right)\right)^N \\ &= \left(1 + \frac{tf'_N(0)}{N}\right)^N = \left(1 + \frac{N}{2}t\right)^N. \end{aligned}$$

Solution 3. Here we solve the problem when $N \ge 1$ is an arbitrary real number. For a real number $a$ let
$$f(x) = \left(\frac{x^{2N} + 1}{2}\right)^{\frac{1}{N}} - a(x - 1)^2 - x.$$
Then $f(1) = 0$,
$$f'(x) = \left(\frac{x^{2N} + 1}{2}\right)^{\frac{1}{N}-1} x^{2N-1} - 2a(x-1) - 1, \quad f'(1) = 0$$
$$f''(x) = (1-N)\left(\frac{x^{2N} + 1}{2}\right)^{\frac{1}{N}-2} x^{4N-2} + (2N-1)\left(\frac{x^{2N} + 1}{2}\right)^{\frac{1}{N}-1} x^{2N-2} - 2a,$$
$$f''(1) = N - 2a.$$
So if $a < \frac{N}{2}$, the function $f$ has a strict local minimum at point 1, and the inequality $f(x) \le 0 = f(1)$ does not hold. This proves $b_N \ge N/2$.
For $a = \frac{N}{2}$ we have $f''(1) = 0$ and
$$f'''(x) = \frac{1}{2}(1-N)(1-2N)\left(\frac{x^{2N} + 1}{2}\right)^{\frac{1}{N}-3}x^{2N} - 3(1-x^{2N}) \begin{cases} > 0, & \text{if } 0 < x < 1, \\ < 0, & \text{if } x > 1. \end{cases}$$
Hence, $f''(x) < 0$ for $x \ne 1$; $f'(x) > 0$ for $x < 1$ and $f'(x) < 0$ for $x > 1$, finally $f(x) < 0$ for $x \ne 1$.