TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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The Cauchy-Schwarz inequality for finite sums

Statement

Let $n \in N$ and let $a_{0}, \dots, a_{n - 1}$ and $b_{0}, \dots, b_{n - 1}$ be reals, with finite sums as in Finite sums and finite products, by recursion. Then

$(\sum_{k < n} a_{k} b_{k})^{2} \leq (\sum_{k < n} a_{k}^{2}) (\sum_{k < n} b_{k}^{2}),$

and, in root form (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ),

$∣ \sum_{k < n} a_{k} b_{k} ∣ \leq \sqrt{\sum_{k < n} a_{k}^{2}} \sqrt{\sum_{k < n} b_{k}^{2}} .$

Equality holds in the first display if and only if the two lists are proportional, in the symmetric sense that there is a pair $(λ, μ) \neq (0, 0)$ of reals with $λ a_{k} = μ b_{k}$ for every $k < n$ .

No root is used in the proof of the squared form. That form is an identity plus a sign argument in the ordered field, and the root form is only a restatement of it through the monotonicity of squaring on the nonnegatives (Squaring is monotone on the nonnegatives); the root enters nowhere but the last step, where that restatement is made and Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ is what supplies the square-root symbol. This matters here, because it makes the squared inequality independent of the existence theorem for roots.

Statement references

Finite sums and finite products, by recursion

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$

Squaring is monotone on the nonnegatives

Facts & Assumptions

Given: A natural $n$ and reals $a_{0}, \dots, a_{n - 1}$ , $b_{0}, \dots, b_{n - 1}$ . Write $A : = \sum_{k < n} a_{k}^{2}$ , $B : = \sum_{k < n} a_{k} b_{k}$ and $C : = \sum_{k < n} b_{k}^{2}$ .

[L1]

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 and vanishes only if every term vanishes.

[L2]

Squares (Squares of nonzero elements are positive, Integer powers $a^{m}$ ): $x^{2} \geq 0$ for every $x$ , and $x^{2} = 0$ only for $x = 0$ ; and a product with a zero factor vanishes, $0 \cdot x = 0$ (Multiplication by zero: $0 \cdot a = 0$ ).

[L3]

Square roots (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ): every $c \geq 0$ has a unique $\sqrt{c} \geq 0$ with $(\sqrt{c})^{2} = c$ .

[L4]

Monotonicity of squaring (Squaring is monotone on the nonnegatives): for $x, y \geq 0$ , $x \leq y ⟺ x^{2} \leq y^{2}$ ; and $∣ x ∣^{2} = x^{2}$ with $∣ x ∣ \geq 0$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L5]

Order arithmetic in an ordered field: adding inequalities (Order is preserved by adding a constant and by adding inequalities) and scaling an inequality by a positive element (Sign rules for products and monotonicity of multiplication, claim 4) are both stated there for the STRICT order alone, so the nonstrict uses below are those statements together with the case of equality, settled by trichotomy (Ordered field); the inverse of a positive element is positive (Inverses of positives are positive, and reciprocation reverses order, claim 1); and a nonzero factor cancels, since $λ x = 0$ with $λ \neq 0$ gives $x = λ^{- 1} (λ x) = λ^{- 1} \cdot 0 = 0$ , a product with a zero factor (Multiplication by zero: $0 \cdot a = 0$ ).

Proof

technique · direct

1.1L1L2

For every $t \in R$ each term $(a_{k} t + b_{k})^{2}$ is nonnegative, so the sum is nonnegative, and expanding with additivity and scaling gives $0 \leq \sum_{k < n} (a_{k} t + b_{k})^{2} = A t^{2} + 2 B t + C$ .

1.2L1L2

In particular $A \geq 0$ and $C \geq 0$ , being sums of squares.

1.3L1L5

Proportionality forces equality: assume $λ a_{k} = μ b_{k}$ for all $k < n$ with $(λ, μ) \neq (0, 0)$ ; if $μ \neq 0$ then $b_{k} = c a_{k}$ with $c = λ / μ$ , so $B = c A$ and $C = c^{2} A$ by scaling, whence $B^{2} = c^{2} A^{2} = A (c^{2} A) = A C$ ; and if $μ = 0$ then $λ \neq 0$ forces $a_{k} = 0$ for all $k$ , so $A = B = 0$ and both sides vanish.

2.1step 1.2L1L2

Suppose first $A = 0$ : then every term of $\sum_{k < n} a_{k}^{2}$ vanishes, so $a_{k} = 0$ for all $k < n$ , hence $B = \sum_{k < n} 0 \cdot b_{k} = 0$ and both sides of the squared inequality are $0$ , so it holds with equality; and the pair $(λ, μ) = (1, 0)$ satisfies $λ a_{k} = 0 = μ b_{k}$ .

2.2step 1.1step 1.2L5

Suppose instead $A > 0$ and substitute $t = - B / A$ into step 1.1: $0 \leq A \frac{B^{2}}{A^{2}} - 2 \frac{B^{2}}{A} + C = C - \frac{B^{2}}{A}$ , so $\frac{B^{2}}{A} \leq C$ , and multiplying by $A > 0$ gives $B^{2} \leq A C$ .

3.1step 2.1step 2.2step 1.2

The squared inequality therefore holds in both cases, which exhaust the possibilities since $A \geq 0$ .

3.2step 1.1step 2.1step 2.2L1L2L5

Equality forces proportionality: in the case $A = 0$ this was step 2.1; in the case $A > 0$ , if $B^{2} = A C$ then putting $t_{0} = - B / A$ in step 1.1 gives $\sum_{k < n} (a_{k} t_{0} + b_{k})^{2} = A t_{0}^{2} + 2 B t_{0} + C = C - \frac{B^{2}}{A} = 0$ , so every term vanishes and $b_{k} = - t_{0} a_{k}$ for all $k$ , and the pair $(λ, μ) = (- t_{0}, 1) \neq (0, 0)$ works.

4.1step 3.1step 1.2L3L4∎

The root form: by step 3.1, $∣ B ∣^{2} = B^{2} \leq A C = (\sqrt{A} \sqrt{C})^{2}$ , and both $∣ B ∣$ and $\sqrt{A} \sqrt{C}$ are nonnegative, so monotonicity of squaring on the nonnegatives gives $∣ B ∣ \leq \sqrt{A} \sqrt{C}$ ; note also that $\sqrt{A} \sqrt{C}$ is the nonnegative square root of $A C$ , by uniqueness.

The Cauchy-Schwarz inequality for finite sums

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources