(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove the Cauchy-Schwarz Inequality.

2. Relevant equations

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

3. The attempt at a solution

Ifxis equal to0, then both sides are equal to 0. Ifxnot equal to0the following applies. The dot product of a vector by itself is always greater or equal to0. 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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# Homework Help: Cauchy-Schwarz Inequality Proof Question

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