TheoremBase

Expanding the nonnegative quantity (x - ty)·(x - ty) with t = (x·y)/(y·y) gives (x·y)^2 ≤ (x·x)(y·y), the case y·y = 0 being trivial, and taking square roots, which are monotone and multiplicative, gives the inequality.

Proof

Each result cited is universally quantified over the data in its own statement.

Write a=x⋅xa=x\cdot x, b=x⋅yb=x\cdot y and c=y⋅yc=y\cdot y, real numbers by The Dot Product on Rn\mathbb{R}^n §dot-product and Iterated Operations and Operations over Finite Sets Take Values in the Underlying Set §segment-sums. By The Dot Product on Rn\mathbb{R}^n Is Symmetric, Bilinear and Positive Definite §nonnegative, a≥0a\ge0 and c≥0c\ge0. By The Euclidean Norm on Rn\mathbb{R}^n §norm and The k-th Root and the Square Root of a Nonnegative Real Number §square-root, ∥x∥=a\lVert x\rVert=\sqrt{a} and ∥y∥=c\lVert y\rVert=\sqrt{c} are the unique nonnegative real numbers whose squares are aa and cc, uniqueness being given by Existence and Uniqueness of Nonnegative k-th Roots of Nonnegative Real Numbers §root. We use that u2=u1+1=u1u1=u uu^{2}=u^{1+1}=u^{1}u^{1}=u\,u for u∈Ru\in\mathbb{R}, by 2=1+12=1+1 (Arithmetic and Order of the Natural Numbers §digits), Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product. Since bb is real, ∣b∣|b| is its absolute value, by Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces §complex-plane.

Step 1: b b≤a cb\,b\le a\,c. Suppose first c=0c=0. Then y=0y=0, the origin, by The Dot Product on Rn\mathbb{R}^n Is Symmetric, Bilinear and Positive Definite §definite, and 0y=00y=0 by Rules for Pointwise Operations on Real-Valued Maps: a Commutative Ring, Scalar Multiples and Evaluation §constant-multiple with I=[n]I=[n], the origin being the constant map 00 of Pointwise Operations on Real-Valued Maps §constants by Euclidean Space Rn\mathbb{R}^n: n-Tuples of Real Numbers with Their Componentwise Operations §origin; so y=0yy=0y and, by The Dot Product on Rn\mathbb{R}^n Is Symmetric, Bilinear and Positive Definite §second and Rules of Arithmetic and Order in an Ordered Field §zero,

b=x⋅(0y)=0 (x⋅y)=0.b=x\cdot(0y)=0\,(x\cdot y)=0 .

Hence b b=0=a⋅0=a cb\,b=0=a\cdot0=a\,c by Rules of Arithmetic and Order in an Ordered Field §zero and commutativity, and b b≤a cb\,b\le a\,c.

Suppose now c≠0c\neq0; then c>0c>0, as c≥0c\ge0, and c−1c^{-1} exists by Negatives, Differences, Reciprocals and Quotients §reciprocal. Let t=b c−1t=b\,c^{-1}, which lies in R\mathbb{R} by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima §quotient, so that t c=b (c−1c)=b⋅1=bt\,c=b\,(c^{-1}c)=b\cdot1=b. The vector v=x−tyv=x-ty lies in Rn\mathbb{R}^{n} by Euclidean Space Rn\mathbb{R}^n: n-Tuples of Real Numbers with Their Componentwise Operations §operations. By The Dot Product on Rn\mathbb{R}^n Is Symmetric, Bilinear and Positive Definite §second and The Dot Product on Rn\mathbb{R}^n Is Symmetric, Bilinear and Positive Definite §symmetric,

x⋅v=x⋅x−x⋅(ty)=a−t (x⋅y)=a−t b,y⋅v=y⋅x−y⋅(ty)=b−t (y⋅y)=b−t c=b−b=0,x\cdot v=x\cdot x-x\cdot(ty)=a-t\,(x\cdot y)=a-t\,b,\qquad y\cdot v=y\cdot x-y\cdot(ty)=b-t\,(y\cdot y)=b-t\,c=b-b=0,

and then, by The Dot Product on Rn\mathbb{R}^n Is Symmetric, Bilinear and Positive Definite §first,

v⋅v=(x−ty)⋅v=x⋅v−(ty)⋅v=(a−t b)−t (y⋅v)=(a−t b)−t⋅0=a−t b,v\cdot v=(x-ty)\cdot v=x\cdot v-(ty)\cdot v=(a-t\,b)-t\,(y\cdot v)=(a-t\,b)-t\cdot0=a-t\,b,

using Rules of Arithmetic and Order in an Ordered Field §zero and −0=0-0=0 (−0=0-0=0 since 0+0=00+0=0, by Negatives, Differences, Reciprocals and Quotients §negative). By The Dot Product on Rn\mathbb{R}^n Is Symmetric, Bilinear and Positive Definite §nonnegative, 0≤v⋅v=a−t b0\le v\cdot v=a-t\,b, so t b≤at\,b\le a by Rules of Arithmetic and Order in an Ordered Field §order-sum. As c>0c>0, Rules of Arithmetic and Order in an Ordered Field §order-product gives (t b) c≤a c(t\,b)\,c\le a\,c, and (t b) c=b (t c)=b b(t\,b)\,c=b\,(t\,c)=b\,b by associativity and commutativity. So b b≤a cb\,b\le a\,c.

Step 2: square roots. By Step 1 and the identity u2=u uu^{2}=u\,u, b2≤a cb^{2}\le a\,c; moreover b2=b b≥0b^{2}=b\,b\ge0 by Rules of Arithmetic and Order in an Ordered Field §squares, so a c≥0a\,c\ge0 as well. Hence, by Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits §abs, Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits §monotone (with k=2k=2, both b2b^{2} and a ca\,c being nonnegative) and Properties of k-th Roots: Inverse to Powers, Absolute Values, Products, Monotonicity, Subadditivity, Differences and Limits §product (with k=2k=2, a,c≥0a,c\ge0),

∣x⋅y∣=∣b∣=b2≤a c=a c=∥x∥ ∥y∥.|x\cdot y|=|b|=\sqrt{b^{2}}\le\sqrt{a\,c}=\sqrt{a}\,\sqrt{c}=\lVert x\rVert\,\lVert y\rVert .

Citations

Loading…

Dependencies

Uses0

Loading…

Comments

Log in to comment.

Loading…