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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Minkowski's inequality for finite sums and real exponent p greater than one

Statement

Let p>1p>1. For real families (ai)i<n(a_i)_{i<n} and (bi)i<n(b_i)_{i<n}, (i<nai+bip)1/p(i<naip)1/p+(i<nbip)1/p.\left(\sum_{i<n}|a_i+b_i|^p\right)^{1/p}\le\left(\sum_{i<n}|a_i|^p\right)^{1/p}+\left(\sum_{i<n}|b_i|^p\right)^{1/p}.

Facts & Assumptions

Given: A natural nn, a real p>1p>1, and real families ai,bia_i,b_i for i<ni<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+vu+v|u+v|\le|u|+|v| (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Basic properties of the absolute value).

Proof

technique · direct
1.1

Let q=p/(p1)q=p/(p-1) and put C=(i<nai+bip)1/pC=(\sum_{i<n}|a_i+b_i|^p)^{1/p}. If C=0C=0, the claim is immediate.

L2L3
1.2

For C>0C>0, multiply ai+bip1|a_i+b_i|^{p-1} by ai+biai+bi|a_i+b_i|\le|a_i|+|b_i| and sum to get Cpaiai+bip1+biai+bip1C^p\le\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 (p1)q=p(p-1)q=p, their common second factor is Cp1C^{p-1}.

step 1.2L1L3
3.1

Thus Cp(A+B)Cp1C^p\le(A+B)C^{p-1}, where A,BA,B are the two right-side norms; dividing by Cp1>0C^{p-1}>0 proves the claim.

step 1.1step 2.1L3

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