Thanks. In your first sentence, do you mean $y_j - x_j > 0$ for all $j$?

@DietrichBurde : I realize that. Why would it be so terrible if the matrix were singular?

@DavidSpeyer : Thanks for all the time you have obviously put into this problem and your partial results. If I solve this problem one way or the other, with a proof or a counterexample, I will edit my question and leave a comment for you, which you will hopefully be notified of in your e-mail.

if your assertion is correct (I have no reason to doubt it, though I don't understand the proof 100%), I am pretty sure that the determinant of the matrix $M$ has the right sign if $\gamma_{n-1}< \mu_1 < \gamma_{n} < \mu_2$ and and $\gamma_n - \mu_1$ is ``really small''. I'd like to extend this to the case where $\gamma_{n-1}< \mu_1 < \gamma_{n} < \mu_2$ and $\gamma_n - \mu_1$ is not necessarily small.

Thanks for the latest revision to your answer. Knowing that the conjecture is false for negative $B$ may be useful. I do not quite understand the "closed form" at the end, in particular the definition of $\gamma_S$ (what is $S$?).

@DavidSpeyer : I added some stuff to the question.

@DavidSpeyer : I haven't computed derivatives of $d(B)$ numerically, just $d(B)$ itself.

@DavidSpeyer : Thanks, that is helpful. Igor Rivin's answer below also looks good. I am going to wait a week or so and accept my favorite answer.

Thanks, that is really impressive. I had never heard of a Cauchy matrix before. By the way, $B \geq 0$ and all the $\gamma$'s are positive and distinct. David Speyer's commment above also looks helpful. I am going to wait a week or so and accept my favorite answer.

@Gerald : You make a convincing argument that any formally real field $\mathbb{F}$ will work. But I would like something more general, that includes Hermitian inner products. Perhaps, we need a field $\mathbb{F}$ with an ordered subfield $\mathbb{F}_1 \subset \mathbb{F}$ and a field automorphism $\phi: \mathbb{F} \to \mathbb{F}$ such that $\mathbf{x}\phi(\mathbf{x}) \in \mathbb{F}_1$ for all $\mathbf{x} \in \mathbb{F}$ and $\mathbf{x}\phi(\mathbf{x}) > 0$ for all $\mathbf{x} \in \mathbb{F} \setminus \{0\}$. Does anyone know if that is good enough? Is it too much?