= 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 gives
$$
\|F(1)\|^2\le\sup_{|z|=R}\|F(z)\|\;\sup_{|z|=1/R}\|F(z)\|.
$$
Now take $F(z)=p(z)^{2^n}$. Even for noncommuting coefficients this is an algebra-valued <holomorphic> <polynomial>. Taking the $2^{-n}$th power gives
$$
\|p(1)^{2^n}\|^{2/2^n}\le
\max_{|z|=R}\|p(z)^{2^n}\|^{1/2^n}
\max_{|z|=1/R}\|p(z)^{2^n}\|^{1/2^n}.
$$
On each <boundary> circle the <continuous> functions $f_n(z)=\|p(z)^{2^n}\|^{1/2^n}$ 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:
$$
\boxed{r_A(p(1))^2\le
\sup_{|z|=R}r_A(p(z))\;\sup_{|z|=1/R}r_A(p(z)).}
$$
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.
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