Using the Cauchy-Schwarz inequality to prove all real values for a, b, and theta

1. Dec 15, 2011

sam0617

1. The problem statement, all variables and given/known data
Use the Cauchy-Schwarz inequality to prove that for all real values of a, b, and theta (which ill denote as θ),
(a cosθ + b sinθ)2 ≤ a2 + b2

2. Relevant equations
so the Cauchy-Schwarz inequality is | < u,v>| ≤ ||u|| ||v||

3. The attempt at a solution

I'm having a difficult time figuring out what is my u and my v. Sorry for the stupid question.

Thank you for any guidance.

Last edited: Dec 15, 2011
2. Dec 15, 2011

micromass

Staff Emeritus
What is <u,v> and ||u|| in $\mathbb{R}^2$??

Can you rewrite the Cauchy-Schwarz inequality in $\mathbb{R}^2$??

3. Dec 15, 2011

sam0617

<u,v> in R2 would be u1v1 + u2v2 I believe..

and ||u|| in R2 is √u2

I'm terribly sorry but I'm not sure what you're trying to get at. I know and I also don't want to be spoon fed the answer.

4. Dec 15, 2011

micromass

Staff Emeritus
Good

No, that would be $\sqrt{u_1^2+u_2^2}$.

Anybody, do you recognize something like $u_1v_1+u_2v_2$ in your OP??

5. Dec 15, 2011

sam0617

I see that a2 + b2 would be
a12a12+b12b12

6. Dec 15, 2011

micromass

Staff Emeritus
Can you write out

$$<u,v>^2\leq \|u\|^2\|v\|^2$$

??

Just write out what <u,v> and ||u|| mean...

7. Dec 15, 2011

sam0617

Sorry, I think you're losing patience with me but I did what you asked and here it is.
Please correct me if I'm incorrect.

<u,v>2≤∥u∥2∥v∥2
= (u1v1+u2
v2)(u1v1+u2
v2) ≤ √(u12+ u22) √(v12 + v22)

8. Dec 15, 2011

sam0617

Well, thanks anyway. I got help from the wonderful people at yahoo answers. Thank you for trying to work with me.