An informal version of my question is "If we have a Pisot number between 1 and 2 of a very large degree, is it true that all its other conjugates are very close to 1 in modulus?"

A more formal version is as follows: let $\beta_n\in(1,2)$ be a convergent sequence of Pisot numbers with $d_n$ being the degree of $\beta_n$ and $\gamma^{(n)}_1,\dots, \gamma^{(n)}_{d_n-1}$ be the Galois conjugates of $\beta_n$ which are different from $\beta_n$. We have then by definition, $|\gamma_k^{(n)}|<1$ for all $n\ge1$ and all $k=1,\dots,d_n-1$. Put
$$
\mu_n=\min_{1\le k\le d_n-1} |\gamma_k^{(n)}|.
$$
**Question.** Is it true that $\mu_n\to1$ as $n\to\infty$?

A colleague of mine believes he has seen this claim somewhere, so this may be a matter of a straightforward reference.

EDIT. For those familiar with the literature, this claim can be proved directly for the so-called *regular Pisot numbers* $\beta=\lim_{n\to\infty}\beta_n$. Unfortunately, there exist other (irregular) Pisot numbers which may or may not accumulate at 2. This class does not appear to exhibit any pattern. So I've been wondering if there is a way to prove this without addressing this issue.

ADDED EXAMPLE. I think it might be helpful to give an example. Let $\beta_n$ be the $n$th multinacci number, i.e., the root of $x^{n+1}-2x^n+1$ lying between 1 and 2. Let $\gamma=\gamma_k^{(n)}$ be one of its conjugates. We have $\gamma^{n+1}=2\gamma^n-1$, whence $$ |\gamma^{-n}|=\left|\frac2{\gamma}-1\right|\le \left|\frac2{\gamma}\right|+1\le5, $$ since $|\gamma|\ge\frac12$. Thus, $|\gamma|\ge 1/\sqrt[n]5\to1$ as $n\to\infty$. This works so well here, because $\beta_n$ is a root of a polynomial with small coefficients and a small number of nonzero coefficients. It is unclear whether this is always the case.