The arithmetic mean, geometric mean inequality

Base case $P(1)$: for a single entry $b_0 \ge 0$ we have $\prod_{k<1} b_k = b_0$ and $B = \frac{1}{1}b_0 = b_0$, so $\prod_{k<1} b_k = b_0 = B^{1}$; equality always holds, and the condition "all entries equal" is automatic for one entry.

Inductive hypothesis: fix $j \in \mathbb{N}$, write $N := N_j$ for the $j$-th doubling natural, and assume $P(N)$.

Reduction to positive entries: if some $b_k = 0$ then $\prod_{k<m} b_k = 0 \le B^{m}$, and equality means $B^{m} = 0$, hence $B = 0$, hence $\sum_{k<m} b_k = 0$, hence every $b_k = 0$, that is, all entries equal; conversely all entries equal to $0$ gives both sides $0$; so $P(m)$ holds for lists with a zero entry, and it is enough to prove $P(m)$ for lists of strictly positive entries.

The two-entry case $P(2)$: for $x, y \ge 0$, $\big(\tfrac{x+y}{2}\big)^{2} - xy = \tfrac{(x+y)^{2} - 4xy}{4} = \big(\tfrac{x-y}{2}\big)^{2} \ge 0$, so $xy \le \big(\tfrac{x+y}{2}\big)^{2}$, with equality exactly when $\big(\tfrac{x-y}{2}\big)^{2} = 0$, that is when $x = y$.

The doubling naturals grow past their index: $N_j \ge 1$ and $N_j > j$ for every $j \in \mathbb{N}$, by induction on $j$ inside $\mathbb{N}$; at $j = 0$, $N_0 = 1 \ge 1$ and $1 > 0$; and if $N_j \ge 1$ and $N_j > j$, then $N_j \ge j + 1$, so $N_{j+1} = N_j + N_j \ge (j+1) + 1 > j + 1$ and $N_{j+1} \ge 1 + 1 \ge 1$. This inequality is between natural numbers throughout; nothing is transported into $\mathbb{R}$.

Doubling step, $P(N) \Rightarrow P(2N)$, where $2N := N + N = N_{j+1}$ is again a natural number: let $a_0, \dots, a_{2N-1} > 0$, put $X := \frac{1}{N}\sum_{k<N} a_k$ and $Y := \frac{1}{N}\sum_{k=N}^{2N-1} a_k$, both positive, and let $A := \frac{1}{2N}\sum_{k<2N} a_k$; splitting the sum at $N$ gives $\sum_{k<2N} a_k = NX + NY$, so $A = \frac{X+Y}{2}$; splitting the product at $N$ and applying the hypothesis to each half gives $\prod_{k<2N} a_k = \big(\prod_{k<N} a_k\big)\big(\prod_{k=N}^{2N-1} a_k\big) \le X^{N} Y^{N} = (XY)^{N}$; and the two-entry case applied to $X, Y$ gives $XY \le \big(\frac{X+Y}{2}\big)^{2} = A^{2}$, whence $(XY)^{N} \le \big(A^{2}\big)^{N} = A^{2N}$ and therefore $\prod_{k<2N} a_k \le A^{2N}$.

Descent step, $P(m+1) \Rightarrow P(m)$ for every natural $m \ge 1$, written in this direction so that no subtraction of naturals is needed: let $b_0, \dots, b_{m-1} > 0$ and put $B := \frac{1}{m}\sum_{k<m} b_k > 0$; define $a_k := b_k$ for $k < m$ and $a_{m} := B$, so that $\sum_{k<m+1} a_k = mB + B = (m+1)B$ and the arithmetic mean of $a_0, \dots, a_{m}$ is exactly $B$; applying $P(m+1)$ gives $\big(\prod_{k<m} b_k\big)B = \prod_{k<m+1} a_k \le B^{m+1}$, and dividing by $B > 0$ gives $\prod_{k<m} b_k \le B^{m}$; moreover equality here holds exactly when $\prod_{k<m+1} a_k = B^{m+1}$, which by the equality clause of $P(m+1)$ holds exactly when all of $a_0, \dots, a_{m}$ are equal, that is when $b_0 = \dots = b_{m-1} = B$, which in turn holds exactly when the $b_k$ are all equal, since a constant list equals its own mean; together with the zero-entry case this is $P(m)$.

Equality in the doubling step: with the notation of the doubling step 2.1, suppose $\prod_{k<2N} a_k = A^{2N}$; then both inequalities in the chain are equalities, so first $(XY)^{N} = \big(A^{2}\big)^{N}$ with $XY, A^{2} > 0$ gives $XY = A^{2}$ by injectivity, hence $X = Y$ by the equality clause of the two-entry case; and second, writing $P_1 = \prod_{k<N} a_k \le X^{N}$ and $P_2 = \prod_{k=N}^{2N-1} a_k \le Y^{N}$ with all four quantities positive, $P_1 < X^{N}$ would give $P_1 P_2 < X^{N} P_2 \le X^{N} Y^{N}$, so $P_1 = X^{N}$ and likewise $P_2 = Y^{N}$, whence by the equality clause of $P(N)$ the first half is constant, equal to its mean $X$, and the second half is constant, equal to $Y = X$; so all $2N$ entries are equal, and conversely entries all equal to $c$ give $\prod = c^{2N}$ and $A = c$.

By the induction principle, $P(N_j)$ holds for every $j \in \mathbb{N}$: the base is $P(N_0) = P(1)$ and the step is the doubling argument together with its equality clause.

Every $n \ge 1$ is reached: by step 1.5, $N_n > n$, so $n \le N_n$ and $N_n = n + i_0$ for some $i_0 \in \mathbb{N}$; now run a further induction on $i$ over the statement $Q(i)$, "for every natural $m \ge 1$ with $m + i = N_n$, $P(m)$ holds"; $Q(0)$ says $P(N_n)$, which is step 4.1; and if $Q(i)$ holds and $m \ge 1$ satisfies $m + (i+1) = N_n$, then $(m+1) + i = N_n$ with $m + 1 \ge 1$, so $P(m+1)$ holds by $Q(i)$ and the descent step turns it into $P(m)$, giving $Q(i+1)$; applying $Q(i_0)$ to $m = n$ yields $P(n)$; hence $\prod_{k<n} a_k \le A^{n}$ with equality exactly when all $a_k$ are equal.

Translation to the root form: $G = \big(\prod_{k<n} a_k\big)^{1/n}$ satisfies $G \ge 0$ and $G^{n} = \prod_{k<n} a_k \le A^{n}$ with $A \ge 0$, so $G \le A$, since $G > A$ would give $G^{n} > A^{n}$; and $G = A$ holds exactly when $G^{n} = A^{n}$, by injectivity of $x \mapsto x^{n}$ on the nonnegatives, hence exactly when all $a_k$ are equal.