Weighted AM-GM inequality with rational weights

Common denominator, by an induction written out rather than asserted. The claim at $j$ is: any rationals $v_0, \dots, v_{j-1}$ admit a natural $N \ge 1$ and integers $p_0, \dots, p_{j-1}$ with $v_k = p_k/N$ for every $k < j$. At $j = 0$ take $N = 1$, there being no $p_k$ to produce. Assume the claim at $j$ and let $v_0, \dots, v_j$ be given: applying it to $v_0, \dots, v_{j-1}$ yields a natural $N' \ge 1$ and integers $p'_k$ with $v_k = p'_k/N'$, and $v_j$ has a representative $v_j = a/b$ with $b > 0$, hence with $b$ a natural $\ge 1$; then $N := N'b$ is a natural $\ge 1$, and the integers $p_k := p'_k b$ for $k < j$ and $p_j := a N'$ satisfy $p_k/N = (p'_k b)/(N' b) = p'_k/N' = v_k$ and $p_j/N = (aN')/(bN') = a/b = v_j$, since $(x, y) \sim (xc, yc)$ for $c \ne 0$. The claim at $j = n$, applied to $w_0, \dots, w_{n-1}$, fixes a natural $N \ge 1$ and integers $m_0, \dots, m_{n-1}$ with $w_k = m_k/N$. Finally each $m_k$ is a natural: $w_k \ge 0$ read on the positive-denominator representative $m_k/N$ gives $m_k \ge 0$ in $\mathbb{Z}$, and a nonnegative integer is the image of a unique natural.

The numerators sum to $\iota(N)$ in $\mathbb{R}$: applying scaling to the real sequence $k \mapsto \iota(w_k)$ multiplies the hypothesis $\sum_{k<n} \iota(w_k) = 1$ by $\iota(N)$ to give $\sum_{k<n} \iota(N)\iota(w_k) = \iota(N)$, and $\iota(N)\iota(w_k) = \iota(N w_k) = \iota(m_k)$ since $\iota$ is multiplicative and $N w_k = m_k$ in $\mathbb{Q}$, so $\sum_{k<n} \iota(m_k) = \iota(N)$.

The partial sums of the numerators, inside $\mathbb{N}$: define $M : \mathbb{N} \to \mathbb{N}$ by recursion, $M_0 := 0$ and $M_{j+1} := M_j + m_j$ for $j < n$ (and $M_{j+1} := M_j$ for $j \ge n$), so each $M_j$ is a natural number and $M_j \le M_{j+1}$; then $\iota(M_j) = \sum_{k<j} \iota(m_k)$ for every $j \le n$, by induction on $j$, since $\iota(M_0) = \iota(0) = 0 = \sum_{k<0} \iota(m_k)$ and $\iota(M_{j+1}) = \iota(M_j) + \iota(m_j) = \sum_{k<j} \iota(m_k) + \iota(m_j) = \sum_{k<j+1} \iota(m_k)$ by additivity of $\iota$ and the recursion clause for finite sums.

Hence $M_n = N$, an identity between natural numbers: steps 2.1 and 2.2 give $\iota(M_n) = \sum_{k<n} \iota(m_k) = \iota(N)$, and $m \mapsto \iota(m)$ is injective on $\mathbb{N}$.

The expanded list: define $c : \mathbb{N} \to \mathbb{R}$ by $c_i := x_j$ whenever $M_j \le i < M_{j+1}$ for some $j < n$, and $c_i := 1$ for $i \ge N$; this covers every $i \in \mathbb{N}$ exactly once, because the blocks $[M_j, M_{j+1})$ for $j < n$ partition $[0, M_n) = [0, N)$ by step 3.1, and every $c_i$ with $i < N$ is positive since each $x_j > 0$.

Its sum and product, by a second induction written out rather than asserted. The claim at $j \le n$ is $\sum_{i<M_j} c_i = \sum_{k<j} \iota(m_k) x_k$ and $\prod_{i<M_j} c_i = \prod_{k<j} x_k^{\,m_k}$. At $j = 0$ we have $M_0 = 0$ and all four expressions are the empty sum $0$ or the empty product $1$. Assume the claim at $j < n$. Since $M_j \le M_{j+1}$, splitting gives $\sum_{i<M_{j+1}} c_i = \sum_{i<M_j} c_i + \sum_{i=M_j}^{M_{j+1}-1} c_i$, and the tail is by definition $\sum_{t<m_j} c_{M_j+t}$, since $M_{j+1} - M_j = m_j$; every one of its terms equals $x_j$, because $M_j \le M_j + t < M_{j+1}$ for $t < m_j$, so the constant sum evaluates it as $\iota(m_j)x_j$, and therefore $\sum_{i<M_{j+1}} c_i = \sum_{k<j} \iota(m_k)x_k + \iota(m_j)x_j = \sum_{k<j+1} \iota(m_k)x_k$ by the recursion clause for finite sums. The same computation with products in place of sums, the constant product in place of the constant sum, gives $\prod_{i<M_{j+1}} c_i = \big(\prod_{k<j} x_k^{\,m_k}\big) x_j^{\,m_j} = \prod_{k<j+1} x_k^{\,m_k}$. At $j = n$, where $M_n = N$ by step 3.1, this reads $\sum_{i<N} c_i = \sum_{j<n} \iota(m_j) x_j$ and $\prod_{i<N} c_i = \prod_{j<n} x_j^{\,m_j}$.

Applying AM-GM to $c_0, \dots, c_{N-1}$ and substituting: $\Big(\prod_{j<n} x_j^{\,m_j}\Big)^{1/N} = \Big(\prod_{i<N} c_i\Big)^{1/N} \le \frac{1}{N}\sum_{i<N} c_i = \iota(N)^{-1}\sum_{j<n} \iota(m_j) x_j = \sum_{j<n} \iota(w_j) x_j$, the last equality by scaling together with $\iota(N)^{-1}\iota(m_j) = \iota(m_j/N) = \iota(w_j)$.

Rewriting the left-hand side with the rational power laws: $\Big(\prod_{j<n} x_j^{\,m_j}\Big)^{1/N} = \prod_{j<n} \Big(x_j^{\,m_j}\Big)^{1/N} = \prod_{j<n} x_j^{\,m_j/N} = \prod_{j<n} x_j^{\,w_j}$, each $x_j^{\,m_j}$ being positive.

Combining the two displays gives $\prod_{j<n} x_j^{\,w_j} \le \sum_{j<n} \iota(w_j) x_j$, which is the assertion.