# Help with Cauchy-Schwartz Inequality proof.

1. Jan 26, 2004

### faust9

Help with Cauchy-Schwarz Inequality proof.

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.

Last edited: Jan 26, 2004
2. Jan 26, 2004

### 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 $$0 \leq \theta \leq 2\pi$$

3. Jan 26, 2004

### 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

Last edited: Jan 26, 2004
4. Jan 26, 2004

### himanshu121

Yes both the prove are right

5. Jan 26, 2004

### faust9

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