Let $\mathcal{E}/\mathbb{Q}(t)$ be given by $$y^2=x^3+A(T)x+B(T)$$ for some $A(T),B(T)\in\mathbb{Q}[T]$ and assume $\mathcal{E}$ is non-isotrivial (the $j$-invariant $\frac{6912 A(T)^3}{4A(T)^3 + 27B(T)^2}$ is not constant). Does there necessarily exist $t\in\mathbb{Q}$ such that the Mordell-Weil group of $\mathcal{E}_t/\mathbb{Q}:y^2=x^3+A(t)x+B(t)$ has positive rank?

An unconditional proof or explicit counterexample would be wonderful, but if that's not possible, I would be okay with a conditional proof assuming standard conjectures (e.g. BSD), or a proof for as wide a class of curves as possible, or a discussion of some properties a hypothetical counterexample would have.

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Here are my thoughts so far:

 - We expect that for "most" families, at least $50\%$ of all specializations (ordering $t$ by height) are positive rank. One result in this direction is by Helfgott, who [shows][1] (assuming some standard conjectures) that if $\mathcal{E}$ has a finite place of multiplicative reduction, then half of the specializations $\mathcal{E}_t$ ordered by height have root number $-1$, and therefore have positive rank assuming the parity conjecture. 
 - In contrast to Helfgott's work, there exist non-isotrivial families of curves with constant root number: for example, $W(\mathcal{E}_t)\equiv -1$ [due to Rizzo][2] and $W(\mathcal{E}_t)\equiv 1$ [due to Bettin, David, and Delaunay][3]. It turns out that in both of these families, $100\%$ of the specializations have positive rank (again assuming the parity conjecture), but it certainly may be possible to have a family with generic rank $0$ and constant root number $1$. Even in this case, though, I would still expect some rank $\geq 2$ specializations.
 - Joe Silverman [conjectures][4] that in fact any non-isotrivial family should have *infinitely many* positive rank specializations, but notes that it's not clear how one would prove this. My question is weaker (I'm only asking for a single specialization), and perhaps naively I would hope this makes it more tractable. 
 - For any particular family, it is often possible to explicitly construct a rank $1$ subfamily (as Joe mentions in the answer I cited above, and Siksek [demonstrates][5]). There likely isn't any way to turn this into a general construction guaranteed to work for all families (if there were, it would prove Joe's conjecture).
 

  [1]: https://arxiv.org/pdf/math/0408141.pdf
  [2]: https://www.cambridge.org/core/journals/compositio-mathematica/article/average-root-numbers-for-a-nonconstant-family-of-elliptic-curves/600186E244EFF6C2F35DF02073358459
  [3]: https://www.archives-ouvertes.fr/hal-01478267/document
  [4]: https://mathoverflow.net/a/63970/404359
  [5]: https://mathoverflow.net/a/63856/404359