Classical separation exponent at least 2

For every real number $\varepsilon>0$ and every natural number $S\ge 2$, there exist a natural number $n$ and a Boolean function $f:\{0,1\}^n\to\{0,1\}$ satisfying $S\le s(f)$ and $s(f)^{\,2-\varepsilon}\le b(f)$, where the exponentiation and this latter comparison are in the real numbers, after viewing the natural numbers $s(f)$ and $b(f)$ as real numbers. Here, for an input $x$ and a set $B$ of coordinate indices, $x^B$ complements exactly the coordinates in $B$; $s(f)$ is the maximum over all inputs $x$ of the number of indices $i$ with $f(x^{\{i\}})\ne f(x)$; and $b(f)$ is the maximum over all inputs $x$ of the largest cardinality of a family of distinct, pairwise disjoint, nonempty sets $B$ for which $f(x^B)\ne f(x)$ for every member. The empty family is permitted and contributes zero. The two maxima need not be attained at the same input. The witnesses $n$ and $f$ may depend on both $\varepsilon$ and $S$; no single function or fixed dimension is required to work for all choices. The quantified dimension formally includes $n=0$, but that dimension has sensitivity zero and cannot satisfy $S\le s(f)$ when $S\ge2$; in fact any witness has $n\ge S$. There is no upper bound on $\varepsilon$: $\varepsilon=2$ and $\varepsilon>2$ are included, producing respectively zero and negative exponents, while $\varepsilon=0$ is excluded. Since a witness has $s(f)\ge2$, the base of the real power is positive. The statement requires the inequality for every positive $\varepsilon$ separately and does not assert $s(f)^2\le b(f)$ for any witness.