 Solution

We first extend the scalar Hadamard three-circle theorem to a holomorphic Banach-space-valued function $F$. Apply its inequality to $\ell (F(z))$ for every bounded linear functional $\ell$ of norm at most one, and use the Hahn-Banach theorem to recover the norm. For the radii $1/R,1,R$ this givesNow take $F(z)=p(z{)}^{2^n}$. Even for noncommuting coefficients this is an algebra-valued holomorphic polynomial. Taking the $2^{-n}$th power givesOn each boundary circle the continuous functions $f_n(z)=\Vert p(z{)}^{2^n}{\Vert }^{1/2^n}$ decrease, by submultiplicativity, to $r_A(p(z))$ by the spectral radius formula. Their maxima decrease to the supremum of this pointwise limit. To justify that passage without assuming continuity of the spectral radius, take maximizing points along a convergent subsequence on the compact circle. For any fixed $k$, all later $f_n$ are bounded by $f_k$; continuity of $f_k$ therefore bounds the limiting maxima by $f_k$ at the limiting point. Taking the infimum over $k$ gives the desired upper bound, while the reverse bound is immediate.

Passing to the limit proves the spectral-radius three-circle inequality at the geometric mean:For a nonunital algebra, use its unitization of an algebra; the spectral radius of an original algebra element is unchanged. Zero boundary suprema cause no difficulty in the decreasing-limit argument.