TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Minkowski's inequality for finite sums and real exponent p greater than one

Statement

Let $p > 1$ . For real families $(a_{i})_{i < n}$ and $(b_{i})_{i < n}$ , ${(\sum_{i < n} ∣ a_{i} + b_{i} ∣^{p})}^{1 / p} \leq {(\sum_{i < n} ∣ a_{i} ∣^{p})}^{1 / p} + {(\sum_{i < n} ∣ b_{i} ∣^{p})}^{1 / p} .$

Facts & Assumptions

Given: A natural $n$ , a real $p > 1$ , and real families $a_{i}, b_{i}$ for $i < n$ .

[L1]

Holder's inequality holds for finite sums and conjugate real exponents (Holder's inequality for finite sums and conjugate real exponents).

[L2]

Finite sums distribute over addition, and $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Basic properties of the absolute value).

[L3]

Positive-base real-power laws hold; the zero-base positive-exponent convention gives $0^{r} = 0$ , and positive-base real powers are positive (Real powers for positive bases, with the zero-base positive-exponent convention, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

Proof

technique · direct

1.1

Let $q = p / (p - 1)$ and put $C = (\sum_{i < n} ∣ a_{i} + b_{i} ∣^{p})^{1 / p}$ . If $C = 0$ , the claim is immediate.

L2L3

1.2

For $C > 0$ , multiply $∣ a_{i} + b_{i} ∣^{p - 1}$ by $∣ a_{i} + b_{i} ∣ \leq ∣ a_{i} ∣ + ∣ b_{i} ∣$ and sum to get $C^{p} \leq \sum ∣ a_{i} ∣ ∣ a_{i} + b_{i} ∣^{p - 1} + \sum ∣ b_{i} ∣ ∣ a_{i} + b_{i} ∣^{p - 1}$ .

L2L3

2.1

Apply Holder to both sums in step 1.2. Since $(p - 1) q = p$ , their common second factor is $C^{p - 1}$ .

step 1.2L1L3

3.1

Thus $C^{p} \leq (A + B) C^{p - 1}$ , where $A, B$ are the two right-side norms; dividing by $C^{p - 1} > 0$ proves the claim.

step 1.1step 2.1L3∎

Depends on

Used by

On a finite counting space, Minkowski agrees with the published finite theorem for p>1 Remark

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)