This follows from Julienne's answer. Fix $\lambda \in (0,1)$ and let $$ f_n(x) = \sum_{k=0}^n {n \choose k} \lambda^{k(n-k)}x ^k. $$ Each $f_n$ satisfies condition (a) for $\mu = 1$. As for (b), we can compute the derivative of $f_n$ to get $$f_n ' (x) = n \lambda^{n-1} f_{n-1} \left(\frac{x}{\lambda}\right).$$ Now use induction on $n$: for $n=1$ we have $f_1(x) = 1+x$, so $x=-1$ is its only root. If $f_{n-1}$ has all roots on the unit circle, then all roots $x_0$ of $f_n'$ satisfy $$n\lambda^{n-1}f_{n-1}\left( \frac{x_0}{\lambda}\right)=0$$ and so $x_0=\lambda z_0$ for some $z_0$ on the unit circle, by the induction hypothesis. Hence, as $\lambda \in (0,1)$, $x_0$ lies inside the unit circle. By induction on $n$, (b) holds for all $f_n$ and so the result follows from Cohn.

(Sorry for not leaving this as a comment to Julienne's answer, I do not have enough reputation to comment.)

This follows from Julienne's answer. Fix $\lambda \in (0,1)$ and let $$ f_n(x) = \sum_{k=0}^n {n \choose k} \lambda^{k(n-k)}x ^k. $$ Each $f_n$ satisfies condition (a) for $\mu = 1$. As for (b), we can compute the derivative of $f_n$ to get $$f_n ' (x) = n \lambda^{n-1} f_{n-1} \left(\frac{x}{\lambda}\right).$$ Now use induction on $n$: for $n=1$ we have $f_1(x) = 1+x$, so $x=-1$ is its only root. If $f_{n-1}$ has all roots on the unit circle, then all roots $x_0$ of $f_n'$ satisfy $$n\lambda^{n-1}f_{n-1}\left( \frac{x_0}{\lambda}\right)=0$$ and so $x_0=\lambda z_0$ for some $z_0$ on the unit circle, by the induction hypothesis. Hence, as $\lambda \in (0,1)$, $x_0$ lies inside the unit circle. By induction on $n$, (b) holds for all $f_n$ and so the result follows from Cohn.

(Sorry for not leaving this as a comment to Julienne's answer, I do not have enough reputation to comment.)

Edit: corrected formula for $f_n'$ thanks to Ian Agol's correction.

This follows from Julienne's answer. Fix $\lambda \in (0,1)$ and let $$ f_n(x) = \sum_{k=0}^n {n \choose k} \lambda^{k(n-k)}x ^k. $$ Each $f_n$ satisfies condition (a) for $\mu = 1$. As for (b), we can compute the derivative of $f_n$ to get $$f_n ' (x) = n \lambda^n f_{n-1} \left(\frac{x}{\lambda}\right).$$ Now use induction on $n$: for $n=1$ we have $f_1(x) = 1+x$, so $x=-1$ is its only root. If $f_{n-1}$ has all roots on the unit circle, then all roots $x_0$ of $f_n'$ satisfy $$n\lambda^nf_{n-1}\left( \frac{x_0}{\lambda}\right)=0$$ and so $x_0=\lambda z_0$ for some $z_0$ on the unit circle, by the induction hypothesis. Hence, as $\lambda \in (0,1)$, $x_0$ lies inside the unit circle. By induction on $n$, (b) holds for all $f_n$ and so the result follows from Cohn.

(Sorry for not leaving this as a comment to Julienne's answer, I do not have enough reputation to comment.)

This follows from Julienne's answer. Fix $\lambda \in (0,1)$ and let $$ f_n(x) = \sum_{k=0}^n {n \choose k} \lambda^{k(n-k)}x ^k. $$ Each $f_n$ satisfies condition (a) for $\mu = 1$. As for (b), we can compute the derivative of $f_n$ to get $$f_n ' (x) = n \lambda^{n-1} f_{n-1} \left(\frac{x}{\lambda}\right).$$ Now use induction on $n$: for $n=1$ we have $f_1(x) = 1+x$, so $x=-1$ is its only root. If $f_{n-1}$ has all roots on the unit circle, then all roots $x_0$ of $f_n'$ satisfy $$n\lambda^{n-1}f_{n-1}\left( \frac{x_0}{\lambda}\right)=0$$ and so $x_0=\lambda z_0$ for some $z_0$ on the unit circle, by the induction hypothesis. Hence, as $\lambda \in (0,1)$, $x_0$ lies inside the unit circle. By induction on $n$, (b) holds for all $f_n$ and so the result follows from Cohn.

(Sorry for not leaving this as a comment to Julienne's answer, I do not have enough reputation to comment.)