Constructing Jacobians of rank 1

Let $K$ be a number field, let $g \geq 1$ be an integer and let $f(x) = (x - a_1) \cdots (x - a_{2g + 1}) \in O_K[x]$ be a polynomial that splits into $2g + 1$ distinct linear factors. Write $C$ for the hyperelliptic curve given by $C: y^2 = f(x)$ and write $J = \mathrm{Jac}(C)$ for its Jacobian. Under mild technical assumptions on $f$ that are satisfied almost always, we prove that there exists some $d \in K^\times$ such that the quadratic twist $J^d$ has rank exactly equal to $1$. As a consequence, we deduce that for any positive integer $g$, there exists an absolutely simple abelian variety over $K$ with dimension equal to $g$ and rank equal to $1$.