Cauchy-Schwarz Inequality Proof Question

  • Thread starter Moridin
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Homework Statement



Prove the Cauchy-Schwarz Inequality.

Homework Equations



[tex]|\mathbf{x \cdot y}| \leq |\mathbf{x}||\mathbf{y}|, \forall \mathbf{x,y} \in \mathbb{R}^{n}[/tex] (1)

The Attempt at a Solution



If x is equal to 0, then both sides are equal to 0. If x not equal to 0 the following applies. The dot product of a vector by itself is always greater or equal to 0. This means that the following holds:

[tex]\mathbf{0} \leq (t \mathbf{x} + \mathbf{y}) \cdot (t \mathbf{x} + \mathbf{y}) = t^{2} |\mathbf{x}|^{2} + 2t\mathbf{x} \cdot \mathbf{y} + |\mathbf{y}|^{2}[/tex] (2)

Factoring |x|2 (which is not equal to zero) gives:

[tex]|\mathbf{x}|^{2} (t^{2} + 2 \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^{2}}t + \frac{|\mathbf{y}|^{2}}{|\mathbf{x}|^{2}})[/tex] (3)

Completing the square gives:

[tex]|\mathbf{x}|^{2} ( (t + \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^{2}})^{2} - \frac{(\mathbf{x} \cdot \mathbf{y})^{2}}{|\mathbf{x}|^{4}} + \frac{|\mathbf{y}|^{2}}{|\mathbf{x}|^{2}})[/tex] (4)

Expansion gives:

[tex]|\mathbf{x}|^{2}(t + \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^{2}})^2 - \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^{2}} + |\mathbf{y}|^{2}[/tex] (5)

The parenthesis is equal to zero for an appropriate value of t, that is:

[tex]t = - \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^2}[/tex] (6)

Then the following is true:

[tex]0 \leq |\mathbf{y}|^{2} - \frac{\mathbf{x} \cdot \mathbf{y}^{2}}{|\mathbf{x}|^{2}}[/tex] (7)

Which, rearranged and the root taken of both sides, becomes:

[tex]|\mathbf{x \cdot y}| \leq |\mathbf{x}||\mathbf{y}|[/tex] (8)

Analogous treatment for y. Q.E.D

My question regarding this attempt is with regards to the justification of moving from step 5 to 6 and 7. Is it valid just to pick a specific t, even though the first step does not fixate it? Or is that t that t which gives the lowest value of the RHS in the inequality, so all other t makes the RHS increase, so that if it is valid for the smallest possible RHS (which is the result of that t), then it is possible for all RHS larger (any other t)? The latter seems more intuitive, since the square cannot possible be negative, so the minimum it can get is 0?

Thank you for your time, have a nice day.
 
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Answers and Replies

  • #2
Dick
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It's true for all t, so what could be wrong with picking a specific t? You could have picked any other t and still gotten a valid inequality, it just wouldn't be Cauchy-Schwarz.
 
  • #3
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Right, since the inequality holds for all possible t, it must hold for any tn that is a possible t.

Thanks.
 

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