How is the algebraic form of the Cauchy-Schwarz inequality for $n=2$ derived from the vector form?

Conditions

• Consider the special case where dimension $n=2$
• Vectors are defined by components $a=(a1,a2)$ and $b=(b1,b2)$

Reasoning, step by step

1. Start with the vector form: |$a\cdot b$| ≤ |a||b|.
2. Substitute component definitions: $a\cdot b = a1b1 + a2b2$.
3. Substitute magnitude definitions: |a| = √(a1²+a2²) and |b| = √(b1²+b2²).
4. Form the inequality: |$a1b1 + a2b2$| ≤ √(a1²+a2²) · √(b1²+b2²).
5. Square both sides to remove square roots and absolute values.
6. Result: ($a1b1 + a2b2$)² ≤ (a1²+a2²)(b1²+b2²).

Example

The script details: 'Taking special case $n=2$... vectors $a=(a1,a2)$ and $b=(b1,b2)$ are defined. Substituting into the previous vector inequality yields |$a1 b1 + a2 b2$| ≤ √($a1^2+a2^2$) · √($b1^2+b2^2$). Squaring both sides recovers the algebraic inequality.'

Common misconceptions

• Forgetting to square both sides after handling the absolute value.
• Confusing the sum of squares inside the root with the square of the sum.

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Connected concepts

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