, 2023

Let $1 \le a < b \le n$ be such that $2^{b-a} x_a x_b$ is maximal. This choice of $a$ and $b$ implies that
$$x_{a+t} \le 2^t x_a, \forall t = 1-a, 2-a, \dots, b-a-1,$$
and similarly
$$x_{b-t} \le 2^t x_b, \forall t = b-n, b-n+1, \dots, b-a+1.$$
Now, suppose that $x_a \in (\frac{1}{2^{u+1}}, \frac{1}{2^u}]$ and $x_b \in (\frac{1}{2^{v+1}}, \frac{1}{2^v}]$, and write $x_a = 2^{-\alpha}$, $x_b = 2^{-\beta}$. Then
$$\sum_{i=1}^{a+u-1} x_i \le 2^u x_a \left( \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{a+u-1}} \right) < 2^u x_a \le 1,$$
and similarly,
$$\sum_{i=b-v+1}^{n} x_i \le 2^v x_b \left( \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n-b+v}} \right) < 2^v x_b \le 1,$$
In other words, the sum of the $x'_i$s for $i$ outside of the interval $[a+u, b-v]$ is strictly less than $2$. Since the total sum is at least $3$, and each term is at most $1$, it follows that this interval must have at least two integers, i.e., $a+u < b-v$. Thus, by bounding the sum of the $x_i$, for $i \in [1, a+u] \cup [b-v, n]$ like above, and trivially bounding each $x_i \in (a+u, b-v)$ by $1$, we obtain
$$\begin{aligned} s &< 2^{u+1} x_a + 2^{v+1} x_b + ((b-v-(a+u)-1) \\ &= b-a + (2^{u+1-\alpha} + 2^{v+1-\beta} - (u+v+1)). \end{aligned}$$
Now recall $\alpha \in (u, u+1]$ and $\beta \in (v, v+1]$, so applying Bernoulli's inequality yields
$$\begin{aligned} 2^{u+1-\alpha} + 2^{v+1-\beta} - u - v - 1 &\le (1+(u+1-\alpha)) + (1+(v+1-\beta)) - u - v - 1 \\ &= 3 - \alpha - \beta. \end{aligned}$$
It follows that $s - 3 < b - a - \alpha - \beta$, and so
$$2^{s-3} < 2^{b-a-\alpha-\beta} = 2^{b-a} x_a x_b.$$