TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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.

Cauchy-Schwarz $\lvert\langle x,y\rangle\rvert \le \lVert x\rVert_2\lVert y\rVert_2$ with its equality case, the triangle inequality for $\lVert\cdot\rVert_2$ , the parallelogram law and polarisation

Statement

Let $n \in \mathbb{N}$ and let $x, y \in \mathbb{R}^{n}$ , with the Euclidean inner product and the Euclidean norm as in The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ . Then:

Cauchy-Schwarz. $\lvert\langle x,y\rangle\rvert \;\le\; \lVert x\rVert_2\,\lVert y\rVert_2 ,$ with equality if and only if there is a pair $(\lambda,\mu) \ne (0,0)$ of reals with $\lambda x_k = \mu y_k$ for every $k < n$ .
$\lVert\cdot\rVert_2$ is a norm on $\mathbb{R}^{n}$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms), for every $n \in \mathbb{N}$ ; the metric it induces is $d_2$ of $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it whenever $n \ge 1$ .
Parallelogram law. $\lVert x+y\rVert_2^{2} + \lVert x-y\rVert_2^{2} \;=\; 2\lVert x\rVert_2^{2} + 2\lVert y\rVert_2^{2} .$
Polarisation. $\langle x,y\rangle \;=\; \tfrac14\Bigl(\lVert x+y\rVert_2^{2} - \lVert x-y\rVert_2^{2}\Bigr),$ so the inner product is recovered from the norm it induces.

Clause 1 is a citation, not a new proof. The inequality and its equality case are the published The Cauchy-Schwarz inequality for finite sums, stated there for two lists of reals; all that happens below is that it is read in the vector notation of The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ . Re-proving it here would put two proofs of one statement in the library.

Facts & Assumptions

Given: A natural number $n$ and vectors $x, y \in \mathbb{R}^{n}$ , so that $\langle x,y\rangle = \sum_{k<n}x_ky_k$ and $\lVert x\rVert_2 = \sqrt{\langle x,x\rangle}$ (The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ , Finite sums and finite products, by recursion).

[L1]

Cauchy-Schwarz for finite sums (The Cauchy-Schwarz inequality for finite sums): $\bigl(\sum_{k<n}a_kb_k\bigr)^{2} \le \bigl(\sum_{k<n}a_k^{2}\bigr)\bigl(\sum_{k<n}b_k^{2}\bigr)$ , with equality if and only if there is $(\lambda,\mu) \ne (0,0)$ with $\lambda a_k = \mu b_k$ for every $k<n$ ; and the root form $\bigl|\sum_{k<n}a_kb_k\bigr| \le \sqrt{\sum_{k<n}a_k^{2}}\sqrt{\sum_{k<n}b_k^{2}}$ .

[L2]

The inner product is symmetric, bilinear and positive definite, $\langle x,x\rangle = \sum_{k<n}x_k^{2} \ge 0$ , and $\langle x,x\rangle = 0$ exactly when $x = 0$ (The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ , Laws of finite sums and finite products).

[L3]

Square roots (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ ): every $c \ge 0$ has a unique $s \ge 0$ with $s^{2} = c$ , written $\sqrt{c}$ ; hence $\lVert x\rVert_2 \ge 0$ and $\lVert x\rVert_2^{2} = \langle x,x\rangle$ (Integer powers $a^m$ ).

[L4]

Squaring is monotone on the nonnegatives: for $a,b \ge 0$ , $a \le b$ if and only if $a^{2} \le b^{2}$ , and $a = b$ if and only if $a^{2} = b^{2}$ (Squaring is monotone on the nonnegatives).

[L5]

Absolute value (Basic properties of the absolute value, Absolute value in an ordered field): $|t| \ge 0$ , $|t|^{2} = t^{2}$ , and $|st| = |s|\,|t|$ .

[L6]

The norm axioms (N1), (N2), (N3) (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms), and the fact that $d_2(u,v) = \lVert u-v\rVert_2$ for $n \ge 1$ ( $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it, The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ clause 6).

Proof

technique · direct

1.1

Instantiating [L1] at $a_k := x_k$ and $b_k := y_k$ gives $\langle x,y\rangle^{2} \le \langle x,x\rangle\,\langle y,y\rangle$ , with equality exactly when some $(\lambda,\mu) \ne (0,0)$ has $\lambda x_k = \mu y_k$ for every $k<n$ .

L1L2

1.2

Both $\lvert\langle x,y\rangle\rvert$ and $\lVert x\rVert_2\lVert y\rVert_2$ are nonnegative, and their squares are $\langle x,y\rangle^{2}$ and $\langle x,x\rangle\langle y,y\rangle$ .

L3L5

1.3

Expanding by bilinearity and symmetry, $\langle x+y, x+y\rangle = \langle x,x\rangle + \langle x,y\rangle + \langle y,x\rangle + \langle y,y\rangle = \lVert x\rVert_2^{2} + 2\langle x,y\rangle + \lVert y\rVert_2^{2}$ .

L2L3

1.4

The same expansion at $x - y = x + (-1)y$ gives $\lVert x-y\rVert_2^{2} = \lVert x\rVert_2^{2} - 2\langle x,y\rangle + \lVert y\rVert_2^{2}$ .

L2L3

1.5

For a scalar $\lambda$ , $\langle \lambda x, \lambda x\rangle = \lambda^{2}\langle x,x\rangle = |\lambda|^{2}\lVert x\rVert_2^{2}$ , so $\lVert \lambda x\rVert_2^{2} = \bigl(|\lambda|\lVert x\rVert_2\bigr)^{2}$ .

L2L3L5

1.6

Axiom (N1) holds: $\lVert x\rVert_2 = 0$ if and only if $\lVert x\rVert_2^{2} = \langle x,x\rangle = 0$ , which by positive definiteness says $x = 0$ .

L2L3L4

2.1

Comparing the squares of step 1.2 through step 1.1 and using monotonicity of squaring on the nonnegatives yields $\lvert\langle x,y\rangle\rvert \le \lVert x\rVert_2\lVert y\rVert_2$ , with equality exactly in the proportional case of step 1.1; this is clause 1.

step 1.1step 1.2L4

2.2

Adding the identities of step 1.3 and step 1.4 gives $\lVert x+y\rVert_2^{2} + \lVert x-y\rVert_2^{2} = 2\lVert x\rVert_2^{2} + 2\lVert y\rVert_2^{2}$ , which is clause 3.

step 1.3step 1.4algebra

2.3

Subtracting the identity of step 1.4 from that of step 1.3 gives $\lVert x+y\rVert_2^{2} - \lVert x-y\rVert_2^{2} = 4\langle x,y\rangle$ , which is clause 4 after dividing by $4$ .

step 1.3step 1.4algebra

2.4

Both $\lVert \lambda x\rVert_2$ and $|\lambda|\lVert x\rVert_2$ are nonnegative and by step 1.5 have equal squares, so $\lVert \lambda x\rVert_2 = |\lambda|\lVert x\rVert_2$ , which is axiom (N2).

step 1.5L3L4L5

3.1

By step 2.1 the middle term of step 1.3 satisfies $2\langle x,y\rangle \le 2\lVert x\rVert_2\lVert y\rVert_2$ , so $\lVert x+y\rVert_2^{2} \le \lVert x\rVert_2^{2} + 2\lVert x\rVert_2\lVert y\rVert_2 + \lVert y\rVert_2^{2} = \bigl(\lVert x\rVert_2 + \lVert y\rVert_2\bigr)^{2}$ .

step 1.3step 2.1L5algebra

4.1

Both $\lVert x+y\rVert_2$ and $\lVert x\rVert_2 + \lVert y\rVert_2$ are nonnegative, so step 3.1 and monotonicity of squaring give $\lVert x+y\rVert_2 \le \lVert x\rVert_2 + \lVert y\rVert_2$ , which is axiom (N3).

step 3.1L3L4

5.1

Steps 2.4, 1.6 and 4.1 are exactly (N1), (N2) and (N3), so $\lVert\cdot\rVert_2$ is a norm on $\mathbb{R}^{n}$ for every $n \in \mathbb{N}$ , and for $n \ge 1$ the metric it induces is $d_2$ ; this is clause 2, and with steps 2.1, 2.2 and 2.3 all four clauses are proved.

step 2.1step 2.2step 2.3step 2.4step 1.6step 4.1L6∎

Remarks

Which route to the triangle inequality was taken. The proof above obtains (N3) by expanding $\lVert x+y\rVert_2^{2}$ and applying Cauchy-Schwarz. The alternative is to quote Minkowski's inequality for finite sums (rational exponent) at the rational exponent $p = 2$ , which states the same inequality directly; that route is equally legitimate and is the one Each $\lVert\cdot\rVert_p$ is a norm on $\mathbb{R}^n$ , and the induced metrics are exactly $d_1$ , $d_2$ and $d_\infty$ of the published metric-spaces page uses for a general exponent. Only one of the two is used here, so no statement is proved twice.
Clause 1 holds at $n = 0$ , where it reads $0 \le 0$ , and the equality case is then satisfied by every pair $(\lambda,\mu) \ne (0,0)$ , the condition quantifying over no indices. Clause 2 also holds at $n = 0$ , the zero space carrying exactly one norm (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). What is not available at $n = 0$ is the metric $d_2$ of $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it, which is why the last sentence of clause 2 carries $n \ge 1$ .
Clauses 3 and 4 are what the companion page uses. The parallelogram law is an identity satisfied by every norm of the form $\sqrt{\langle\cdot,\cdot\rangle}$ , so a norm violating it is not of that form; that is how the companion page rules out $\lVert\cdot\rVert_1$ on $\mathbb{R}^{2}$ . Polarisation says the inner product carries no information the norm does not.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 119 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

Cauchy-Schwarz inequality (Wikipedia) (standard reference, not scraped)
Parallelogram law (Wikipedia) (standard reference, not scraped)
Polarization identity (Wikipedia) (standard reference, not scraped)
J. Demmel, MA221 Lecture 3: Vector Norms (standard reference, not scraped)
G. Zitelli, Math 641 Functional Analysis, Part I (standard reference, not scraped)

Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2\lvert\langle x,y\rangle\rvert \le \lVert x\rVert_2\lVert y\rVert_2∣⟨x,y⟩∣≤∥x∥2​∥y∥2​ with its equality case, the triangle inequality for ∥⋅∥2\lVert\cdot\rVert_2∥⋅∥2​, the parallelogram law and polarisation