SAUDI ARABIAN IMO Booklet 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 $$\sum _{i=1}^{a+u-1}{x}_i\le 2^u{x}_a\left(\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{a+u-1}}\right)<2^u{x}_a\le 1,$$ and similarly, $$\sum _{i=b-v+1}^n{x}_i\le 2^v{x}_b\left(\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{n-b+v}}\right)<2^v{x}_b\le 1,$$ In other words, the sum of the $x_i^{\prime }$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 $$\begin{array}{rl}s& <2^{u+1}{x}_a+2^{v+1}{x}_b+((b-v-(a+u)-1)\\ & =b-a+(2^{u+1-\alpha }+2^{v+1-\beta }-(u+v+1)).\end{array}$$ Now recall $\alpha \in (u,u+1]$ and $\beta \in (v,v+1]$, so applying Bernoulli's inequality yields $$\begin{array}{rl}{2}^{u+1-\alpha }+2^{v+1-\beta }-u-v-1& \le (1+(u+1-\alpha ))+(1+(v+1-\beta ))-u-v-1\\ & =3-\alpha -\beta .\end{array}$$ It follows that $s-3<b-a-\alpha -\beta$, and so $$2^{s-3}<2^{b-a-\alpha -\beta }=2^{b-a}{x}_a{x}_b.$$