Help with Cauchy-Schwartz Inequality proof.

[faust9]

Argh!!!

I've been playing around with this and I can't get it...

Here's what I have thus far:

Given \mid u \cdot v \mid \leq \parallel u \parallel \parallel v \parallel

(u \cdot v)^2 \leq (u_1^2 + u_2^2)(v_1^2 + v_2^2)

(u_1v_1+u_2v_2)^2 \leq (u_1^2 + u_2^2)(v_1^2 + v_2^2)

(u_1v_1+u_2v_2)^2 \leq (u_1^2 + u_2^2)(v_1^2 + v_2^2)

u_1^2v_1^2+u_2^2v_2^2+2u_1v_1u_2v_2 \leq u_1^2v_1^2 + u_2^2v_1^2+u_1^2v_2^2 + u_2^2v_2^2

2u_1v_1u_2v_2 \leq u_2^2v_1^2+u_1^2v_2^2

This is where I get stumped which means I messed up somewhere earlier in my proof. Any help here would be greatly appreciated.

Thanks a lot.

[faust9]

Ok, Try number 2. Does this look right?

given: \mid u \cdot v \mid \leq \parallel u \parallel \parallel v \parallel

\frac{\mid u \cdot v \mid}{\parallel u \parallel \parallel v \parallel} \leq 1

since \cos \theta = \frac{\mid u \cdot v \mid}{\parallel u \parallel \parallel v \parallel} \leq 1

\Rightarrow \cos \theta \leq 1

thus the proof is true because by definition, \cos \theta \leq 1 for all values 0 \leq \theta \leq 2\pi

[himanshu121]

This is a perfect square and would be >=0

-2u_1v_1u_2v_2 + u_2^2v_1^2+u_1^2v_2^2 \geq 0

Now u can proceed

[himanshu121]

Yes both the prove are right

[faust9]

I didn't notice the perfect square... Thanks.