TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Minkowski's inequality for finite sums (rational exponent)

Statement

Let $n \in N$ , let $a_{0}, \dots, a_{n - 1}$ and $b_{0}, \dots, b_{n - 1}$ be reals, and let $p \in Q$ with $p \geq 1$ (Order on the rationals). Then

$(\sum_{k < n} ∣ a_{k} + b_{k} ∣^{p})^{1 / p} \leq (\sum_{k < n} ∣ a_{k} ∣^{p})^{1 / p} + (\sum_{k < n} ∣ b_{k} ∣^{p})^{1 / p} .$

All exponents appearing are positive rationals, so every power is defined for a nonnegative base by Rational powers $a^{r}$ of a positive base and its supplementary clause $0^{r} = 0$ .

The conjugate exponent is rational exactly because $p$ is. For $p > 1$ the proof runs through Hölder with $q = p / (p - 1)$ , and a quotient of rationals with nonzero denominator is a rational (Arithmetic on the rationals). Had $p$ been an arbitrary real, $q$ would still be a real, but $∣ a_{k} ∣^{p}$ would already be undefined: the whole statement lives inside the rational exponents built on this page, as the closing remark of the page explains.

On the case $p = 1$ . It reads $\sum_{k < n} ∣ a_{k} + b_{k} ∣ \leq \sum_{k < n} ∣ a_{k} ∣ + \sum_{k < n} ∣ b_{k} ∣$ and follows by summing the two-term triangle inequality (The triangle inequality) termwise. It is not literally the finite-sum triangle inequality Triangle inequality for finite sums, which compares $∣ \sum_{k < n} c_{k} ∣$ with $\sum_{k < n} ∣ c_{k} ∣$ for one list; combining the two at $p = 1$ gives the familiar $∣ \sum_{k < n} (a_{k} + b_{k}) ∣ \leq \sum_{k < n} ∣ a_{k} ∣ + \sum_{k < n} ∣ b_{k} ∣$ .

Facts & Assumptions

Given: A natural $n$ , reals $a_{0}, \dots, a_{n - 1}$ and $b_{0}, \dots, b_{n - 1}$ , and a rational $p \geq 1$ . Write $S : = \sum_{k < n} ∣ a_{k} + b_{k} ∣^{p}$ , $U : = (\sum_{k < n} ∣ a_{k} ∣^{p})^{1 / p}$ , $V : = (\sum_{k < n} ∣ b_{k} ∣^{p})^{1 / p}$ , and $c_{k} : = ∣ a_{k} + b_{k} ∣^{p - 1}$ when $p > 1$ .

[L1]

Hölder's inequality (Hölder's inequality for finite sums (rational exponents)): for conjugate rationals $p, q > 1$ , $\sum_{k < n} ∣ x_{k} y_{k} ∣ \leq (\sum_{k < n} ∣ x_{k} ∣^{p})^{1 / p} (\sum_{k < n} ∣ y_{k} ∣^{q})^{1 / q}$ .

[L2]

Laws of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, monotonicity, and the fact that a sum of nonnegative terms is nonnegative.

[L3]

Rational power laws (Laws of rational exponents, Rational powers $a^{r}$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ): for $u > 0$ and rationals $r, s$ : $u^{r} > 0$ , $u^{r + s} = u^{r} u^{s}$ , $(u^{r})^{s} = u^{r s}$ , $u^{1} = u$ and $u^{- r} = (u^{r})^{- 1}$ ; and $0^{r} = 0$ for every rational $r > 0$ .

[L4]

Absolute values (Basic properties of the absolute value, Absolute value in an ordered field, The triangle inequality): $∣ x ∣ \geq 0$ , $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , and $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ .

[L5]

Rational arithmetic (Arithmetic on the rationals, The rationals as equivalence classes of pairs of integers, Order on the rationals), carried out in the totally ordered field $Q$ (The rationals form a totally ordered field, which is what makes the order comparisons below legitimate, and which supplies totality, compatibility with addition and closure of the positives under multiplication but NOT $1 > 0$ ; that is The multiplicative identity is positive, valid in $Q$ because $Q$ is an ordered field): for rational $p > 1$ one has $p - 1 > 0$ and, since $1 > 0$ , also $p > p - 1 > 0$ , so the number $q : = p / (p - 1)$ is a rational with $q > 1$ , $\frac{1}{p} + \frac{1}{q} = 1$ , $(p - 1) q = p$ and $1 - \frac{1}{q} = \frac{1}{p}$ .

[L6]

Order arithmetic: Order is preserved by adding a constant and by adding inequalities and Sign rules for products and monotonicity of multiplication state adding inequalities and scaling by a positive element for the STRICT order only, so the nonstrict forms used below (adding two $\leq$ , and scaling a $\leq$ by a nonnegative element) are those statements together with the case of equality, which is settled by trichotomy (Ordered field); and the inverse of a positive element is positive (Inverses of positives are positive, and reciprocation reverses order, claim 1).

Proof

technique · direct

1.1L2L3L4

Every quantity is defined and nonnegative: $∣ a_{k} + b_{k} ∣^{p}$ , $∣ a_{k} ∣^{p}$ and $∣ b_{k} ∣^{p}$ are nonnegative because $p > 0$ , hence so are $S$ , $U$ and $V$ .

1.2L2L3L4

The case $p = 1$ : summing the two-term triangle inequality termwise and using monotonicity and additivity gives $\sum_{k < n} ∣ a_{k} + b_{k} ∣ \leq \sum_{k < n} (∣ a_{k} ∣ + ∣ b_{k} ∣) = \sum_{k < n} ∣ a_{k} ∣ + \sum_{k < n} ∣ b_{k} ∣$ , and since $x^{1 / 1} = x$ this is exactly the assertion at $p = 1$ .

1.3L2L3

The case $S = 0$ : the left-hand side is $0^{1 / p} = 0$ , which is at most the nonnegative right-hand side.

2.1step 1.1step 1.2step 1.3L5

Assume from now on $p > 1$ and $S > 0$ , and put $q : = p / (p - 1)$ , a rational with $q > 1$ conjugate to $p$ , so that $(p - 1) q = p$ and $1 - \frac{1}{q} = \frac{1}{p}$ .

3.1step 2.1L3L4L6

Splitting each term: for $u \geq 0$ one has $u^{p} = u^{1 + (p - 1)} = u \cdot u^{p - 1}$ , valid for $u > 0$ by the addition law and for $u = 0$ because both sides are $0$ ; applying this with $u = ∣ a_{k} + b_{k} ∣$ and then the triangle inequality, multiplied by the nonnegative factor $c_{k} = ∣ a_{k} + b_{k} ∣^{p - 1}$ , gives $∣ a_{k} + b_{k} ∣^{p} \leq (∣ a_{k} ∣ + ∣ b_{k} ∣) c_{k} = ∣ a_{k} ∣ c_{k} + ∣ b_{k} ∣ c_{k}$ for every $k < n$ .

3.2step 2.1L3L4

The auxiliary list has $q$ -th power sum $S$ : $c_{k}^{q} = (∣ a_{k} + b_{k} ∣^{p - 1})^{q} = ∣ a_{k} + b_{k} ∣^{(p - 1) q} = ∣ a_{k} + b_{k} ∣^{p}$ for $∣ a_{k} + b_{k} ∣ > 0$ by the iterated-power law, and both sides are $0$ when $a_{k} + b_{k} = 0$ ; hence $\sum_{k < n} ∣ c_{k} ∣^{q} = S$ .

4.1step 3.1L2

Summing the termwise bound: $S \leq \sum_{k < n} ∣ a_{k} ∣ c_{k} + \sum_{k < n} ∣ b_{k} ∣ c_{k}$ .

4.2step 3.2step 2.1L1L4

Applying Hölder to the pairs $(a_{k}, c_{k})$ and to $(b_{k}, c_{k})$ , and using $∣ a_{k} c_{k} ∣ = ∣ a_{k} ∣ c_{k}$ since $c_{k} \geq 0$ : $\sum_{k < n} ∣ a_{k} ∣ c_{k} \leq U S^{1 / q}$ and $\sum_{k < n} ∣ b_{k} ∣ c_{k} \leq V S^{1 / q}$ .

5.1step 4.1step 4.2L6

Combining, $S \leq (U + V) S^{1 / q}$ .

6.1step 5.1step 2.1step 1.2step 1.3L3L5L6∎

Dividing by $S^{1 / q} > 0$ , which is legitimate because $S > 0$ , and computing $S (S^{1 / q})^{- 1} = S^{1} S^{- 1 / q} = S^{1 - 1 / q} = S^{1 / p}$ , we obtain $S^{1 / p} \leq U + V$ ; together with the case $p = 1$ and the case $S = 0$ this proves the inequality for every rational $p \geq 1$ .

Minkowski's inequality for finite sums (rational exponent)

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources