How does the geometric projection argument prove that |a·b| ≤ |a||b| for non-zero vectors?

How does the geometric projection argument prove that | $a\cdot b$ | ≤ |a||b| for non-zero vectors?

The proof relies on the property that the length of a vector's projection onto another vector cannot exceed the original vector's magnitude. By defining the projection of b onto a as proj_a b = (

a\cdot b

/ |a|^2)a, its magnitude is calculated as |proj_a b| = |

a\cdot b

|/|a|. Since |proj_a b| ≤ |b|, substituting yields |

a\cdot b

|/|a| ≤ |b|, which simplifies to |

a\cdot b

| ≤ |a||b|.

Conditions

Vectors a and b are in Euclidean space
|a| ≠ 0

Reasoning, step by step

Define the dot product as $a\cdot b$ = |a||b|cosθ.
Express the projection of b onto a: proj_a b = (( $a\cdot b$ )/|a|^2)a.
Calculate the magnitude of this projection: |proj_a b| = | $a\cdot b$ |/|a|.
Apply the geometric constraint that projection length ≤ original vector length: |proj_a b| ≤ |b|.
Combine inequalities: | $a\cdot b$ |/|a| ≤ |b|.
Multiply both sides by |a| to derive | $a\cdot b$ | ≤ |a||b|.

Example

The script states: 'first the dot product definition $a\cdot b$ = |a||b|cosθ; then the projection formula proj_a b = ( $a\cdot b$ / |a|^2)a; followed by the magnitude of projection |proj_a b| = | $a\cdot b$ |/|a| ≤ |b|. This leads to the conclusion | $a\cdot b$ | ≤ |a||b|.'