TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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Minkowski's inequality for finite sums (rational exponent)

Statement

Let $n \in \mathbb{N}$ , let $a_0, \dots, a_{n-1}$ and $b_0, \dots, b_{n-1}$ be reals, and let $p \in \mathbb{Q}$ with $p \ge 1$ (Order on the rationals). Then

$\Big(\sum_{k<n}|a_k + b_k|^{p}\Big)^{1/p} \;\le\; \Big(\sum_{k<n}|a_k|^{p}\Big)^{1/p} + \Big(\sum_{k<n}|b_k|^{p}\Big)^{1/p}.$

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 $\sum_{k<n}|a_k+b_k| \le \sum_{k<n}|a_k| + \sum_{k<n}|b_k|$ 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 $\big|\sum_{k<n} c_k\big|$ with $\sum_{k<n}|c_k|$ for one list; combining the two at $p = 1$ gives the familiar $\big|\sum_{k<n}(a_k+b_k)\big| \le \sum_{k<n}|a_k| + \sum_{k<n}|b_k|$ .

Facts & Assumptions

Given: A natural $n$ , reals $a_0, \dots, a_{n-1}$ and $b_0, \dots, b_{n-1}$ , and a rational $p \ge 1$ . Write $S := \sum_{k<n}|a_k+b_k|^{p}$ , $U := \big(\sum_{k<n}|a_k|^{p}\big)^{1/p}$ , $V := \big(\sum_{k<n}|b_k|^{p}\big)^{1/p}$ , 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 $p, q > 1$ , $\sum_{k<n}|x_k y_k| \le \big(\sum_{k<n}|x_k|^{p}\big)^{1/p}\big(\sum_{k<n}|y_k|^{q}\big)^{1/q}$ .

[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 $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ ): for $u > 0$ and rationals $r, s$ : $u^{r} > 0$ , $u^{r+s} = u^{r}u^{s}$ , $\big(u^{r}\big)^{s} = u^{rs}$ , $u^{1} = u$ and $u^{-r} = \big(u^{r}\big)^{-1}$ ; 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): $|x| \ge 0$ , $|xy| = |x||y|$ , and $|x+y| \le |x| + |y|$ .

[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 $\mathbb{Q}$ (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 $\mathbb{Q}$ because $\mathbb{Q}$ 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 $q > 1$ , $\frac1p + \frac1q = 1$ , $(p-1)q = p$ and $1 - \frac1q = \frac1p$ .

[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 $\le$ , and scaling a $\le$ 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

1.1

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 $V$ .

L2L3L4

1.2

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

L2L3L4

1.3

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

L2L3

2.1

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 $1 - \frac1q = \frac1p$ .

step 1.1step 1.2step 1.3L5

3.1

Splitting each term: for $u \ge 0$ 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 $|a_k+b_k|^{p} \le \big(|a_k| + |b_k|\big) c_k = |a_k| c_k + |b_k| c_k$ for every $k < n$ .

step 2.1L3L4L6

3.2

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

step 2.1L3L4

4.1

Summing the termwise bound: $S \le \sum_{k<n}|a_k| c_k + \sum_{k<n}|b_k| c_k$ .

step 3.1L2

4.2

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 $c_k \ge 0$ : $\sum_{k<n}|a_k| c_k \le U\,S^{1/q}$ and $\sum_{k<n}|b_k| c_k \le V\,S^{1/q}$ .

step 3.2step 2.1L1L4

5.1

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

step 4.1step 4.2L6

6.1

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

step 5.1step 2.1step 1.2step 1.3L3L5L6∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 78 results over 30 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)
Finite inequalities (Cornell University) (standard reference, not scraped)
Young, Hölder, and Minkowski inequalities (Oregon State University) (standard reference, not scraped)
Minkowski inequality (Wikipedia) (standard reference, not scraped)
Hölder's inequality (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)