Inner Product Proof: Proving Sums with Algebra and Inner Product Concepts

[Dick]

Why don't you just do it like this?

Step 1: Prove that \langle x,y\rangle=\sum_n x_n y_n defines an inner product.
Step 2: Show that this definition turns the Cauchy-Schwartz inequality into

\left(\sum_j x_j y_j\right)^2\leq\left(\sum_j x_j^2\right)\left(\sum_j y_j^2\right)

Step 3: Make a specific choice of x_j and y_j that turns the inequality into the one you want, and explain why you're allowed to do that.

Steps 1 and 2 are probably unnecessary. We are doing standard inner product, standard Cauchy-Schwarz. It's step 3 that seems to be the obstacle for these 30 posts. Good luck.

 

[Fredrik]

The beginning may be a bit confusing, but it gets better later on.

It didn't get much better, and let's be realistic, no one is going to read that far anyway unless you can make more sense in the beginning.

If you're trying to prove the Cauchy-Schwartz inequality, the standard trick is to note that \langle x,x\rangle\geq 0 for all x, and that this means that

0\leq\langle x+ty,x+ty\rangle

for all vectors x,y and all scalars t. You get the Cauchy-Schwartz inequality by using the properties of the inner product and choosing t to make the right-hand side as small as possible. This is the trick that all the books use, which makes me think that it's the easiest method by far. See Wikipedia for more details.

 

[Fredrik]

Steps 1 and 2 are probably unnecessary. We are doing standard inner product, standard Cauchy-Schwarz. It's step 3 that seems to be the obstacle for these 30 posts.

I guess that explains your frustration.

 

[evilpostingmong]

I read the proof, and it involves the quadratic formula. Dick's right, it does
have a weird trick that comes from nowhere (but makes sense anyway).
Seems kinda advanced, but understandable. I don't know why it was included as
problem #3 in my book.
Thanks for the help guys!