Proof of Cauchy-Shwartz Inequality

1. Dec 12, 2007

smoothman

How would you prove the following:

let V be an inner product space. For v,w $\epsilon$ V we have:

|<v,w>| $\leq$ ||v|| ||w||

with equality if and only if v and w are linearly dependent.

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So far I know that the Cauchy-Schwartz inequality says |< v,w>| is less than or equal to ||v||||w|| for any two vectors v and w in an inner product space.

I also realise i have to prove

|< v, w>|< ||v||||w||

that is, that there is a strict inequality, except in the case where one vector is a multiple of the other.
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any ideas on the proof for this please
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 12, 2007

Moridin

Separate them into two cases, one where vector x = 0, and one where it does not. The first is simple to prove, the starting point for the other case is that the scalar product of a vector with itself is greater or equal to 0 (or is it? Why? Try to understand why this is the case!).

x,y are vectors in [itex]R^n[/tex]; t is an arbitrary scalar.

? $$\leq (tx + y) \cdot (tx + y)$$ = ?...