, 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 $x_{a+t} \le 2^t x_a$ for all $1-a \le t \le b-a-1$, and similarly $x_{b-t} \le 2^t x_b$ for all $b-n \le t \le b-a+1$. 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,
$$\sum_{i=b-v+1}^{n} x_i \le 2^v x_b \left( \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n-b+v}} \right) < 2^u x_b \le 1,$$
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 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
$$s < 2^{u+1}x_a + 2^{v+1}x_b + ((b-v)-(a+u)-1) = b-a+(2^{u+1-\alpha}+2^{v+1-\beta}-(u+v+1)).$$

As in the original solution, we arrive at
$$3 < 2^{u+1}x_a + 2^{v+1}x_b + ((b-v)-(a+u)-1) = b-a+(2^{u+1-\alpha}+2^{v+1-\beta}-(u+v+1)).$$
Now $2^{b-a}x_a x_b \ge 2^{b-a}2^{-u-1}2^{-v-1}$, so it is enough to show $-2 < b - a - u - v - 2$, or $3 < b - a - u - v + 3$. The fact that $u+1-\alpha < 1$ and $v+1-\beta < 1$ implies $2^{u+1-\alpha} + 2^{v+1-\beta} < 4$, and so $3 < b - a + (2 + 2 - u - v - 1) = b - a - u - v - 3$.

Now recall $\alpha \in (u, u+1]$ and $\beta \in (v, v+1]$, so applying Bernoulli's inequality yields
$$2^{u+1-\alpha} + 2^{v+1-\beta} - u - v - 1 \le (1+(u+1-\alpha))+(1+(v+1-\beta)) - u - v - 1 = 3 - \alpha - \beta.$$
The above shows that $c = 1$ is the best possible. A somewhat simpler ending to the proof can be given for $c = \frac{1}{4}$.

**End of solution for $c = \frac{1}{4}$**.