Proof of the Cauchy-Shwarz Inequality

[Jow]

1. Another approach to the proof of the Cauchy-Shwarz Inequality is suggested by figure 16 (sorry, I don't have the image), which shows that, in ℝ2 or ℝ3, llproj$_u{}$vll ≤ llvll. Show that this inequality is equivalent to the Cauchy-Schwartz Inequality.


2. Cauchy-Schawrtz Inequality: lu • vl ≤ llull llvll



3. I substituted proj$_u{}$v with (u•v/u•u)u but I can't think of what to do next.

 

[Jow]

Actually, I think I got it (please tell me if I am correct):

ll(u•v/u•u)ull ≤ llvll
lu•v/u•ul llull ≤ llvll
lu•v/u•ul^2 llull^2 ≤ llvll^2
lu•v/u•ul lu•v/u•ul (u•u) ≤ v•v
(lu•vl*lu•vl)/(u•u) ≤ v•v
lu•vl^2 ≤ (v•v)(u•u)
lu•vl^2 ≤ llvll^2 llull^2
lu•vl ≤ llvll llull

 

[Dick]

Actually, I think I got it (please tell me if I am correct):

ll(u•v/u•u)ull ≤ llvll
lu•v/u•ul llull ≤ llvll
lu•v/u•ul^2 llull^2 ≤ llvll^2
lu•v/u•ul lu•v/u•ul (u•u) ≤ v•v
(lu•vl*lu•vl)/(u•u) ≤ v•v
lu•vl^2 ≤ (v•v)(u•u)
lu•vl^2 ≤ llvll^2 llull^2
lu•vl ≤ llvll llull

That's pretty hard to read. But it look ok to me. There's probably some extra steps in there you don't need. u.u=||u||^2, right?

 

[Jow]

Yeah, I put few extra steps that weren't necessary. I know its difficult to read; I am not really familiar with how to properly input some of the symbols.

 

[LCKurtz]

ll(u•v/u•u)ull ≤ llvll

Yeah, I put few extra steps that weren't necessary. I know its difficult to read; I am not really familiar with how to properly input some of the symbols.

It isn't that difficult

$\| (\frac{u\cdot v}{u\cdot u}) u \| \le \| v \|$
Right click on it to see the code. Put ## at the beginning and end of the code.