Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated
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.

Relaxed inequalities for squared Euclidean norms

Statement

For vectors u,v∈Rd and a>0, ∥u+v∥22≤(1+a)∥u∥22+(1+a−1)∥v∥22,2∣⟨u,v⟩∣≤a∥u∥22+a−1∥v∥22. For m≥1 vectors u1,…,um∈Rd, ∥∑j=1muj∥22≤m∑j=1m∥uj∥22.

Facts & Assumptions

Given: The standard Euclidean inner product and norm on Rd, with ∥x∥22=⟨x,x⟩.

[A1]

The real inner product is symmetric and bilinear, and squared norms are nonnegative.

Proof

1.1givenA1algebra

Expanding the first inequality's right side minus its left side gives [given] a∥u∥22+a−1∥v∥22−2⟨u,v⟩=∥a u−a−1/2v∥22≥0. Replacing v by −v gives the same bound with +2⟨u,v⟩; together these two bounds yield both displayed two-vector inequalities.

1.2A1algebra∎

Expanding each squared difference and collecting diagonal and cross terms [given] gives m∑j=1m∥uj∥22−∥∑j=1muj∥22=∑1≤i<j≤m∥ui−uj∥22≥0. This proves the finite-sum inequality, including m=1, when the right difference is zero.

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources