# Showing Relations

1. Feb 5, 2005

Hello all

Show that the equality signs in Schwarz's Inequality holds if, and only if, the a's and b's are proportional; that is; $$ca_{v} + db_{v} = 0$$ for all v's where c and d are independent of v and not both zero. How would I even begin this? I know Schwarz's Inequality is:

$$(a_1b_1 + a_2b_2 + ... + a_nb_n)^2 \leq (a_1^2 +... + a_n^2)(b_1^2 + ...+ b_n^2)$$ Now we need to show that the equality sign holds given the above conditions.

Would i use the fact of direct proportionality, $$y = kx$$?

Thanks

Last edited: Feb 5, 2005
2. Feb 5, 2005

I am not really sure what $$ca_v + db_v = 0$$ is implying.

3. Feb 5, 2005

### Galileo

Use:
$$\vec a \cdot \vec b = |\vec a||\vec b|\cos \theta$$

4. Feb 5, 2005

i have not really studied vectors help. could you please elaborate further?

Thanks

5. Feb 5, 2005

### Townsend

I don't understand this at all. I have been studying the Cauchy inequality for about a month and I have yet to run across anything that says for equality to hold that a and b must be proportional.

Lets see if we can come up with some instances where equality would hold.

First of all a more general form of the Cauchy inequality is

$$|a_1b_1+a_2b_2+...+a_nb_n| \leq {(a_1^2+a_2^2+...+a_n^2)}^{\frac{1}{2}}{(b_1^2+b_2^2+...+b_n^2)}^{\frac{1}{2}}$$

Now for n=1 equality holds but what about n=2?

$$|a_1b_1+a_2b_2| \leq \sqrt{a_1^2+a_2^2} \sqrt{b_1^2+b_2^2}$$

$${(a_1b_1+a_2b_2)}^2 \leq (a_1^2+a_2^2)(b_1^2+b_2^2)$$

If you expand both sides of the inequality you have

$$a_1^2b_1^2+2a_1b_1a_2b_2+a_2^2b_2^2 \leq a_1^2b_1^2+a_1^2b_2^2+a_2^2b_1^2+a_2^2b_2^2$$

Which can be simplified to

$$2a_1b_2a_2b_1 \leq (a_1b_2)^2+(a_2b_1)^2$$

Which we can see the second degree polynomial there and so we move the RHS to the LHS and we have

$$0 \leq (a_1b_2)^2-2a_1b_2a_2b_1+(a_2b_1)^2$$

And the RHS factors into a perfect square which is clearly always either greater than or equal to 0. But we are interested in when it is equal to 0. We want to know under what conditions the RHS is 0. So we must solve

$$(a_1b_2-a_2b_1)^2=0 \\\\\ \mbox{ so we just solve } \\\\ a_1b_2=a_2b_1$$

Now that might be something. I will work on this some more because I think it is very interesting. But for right now I need to run...

Best of luck

Last edited: Feb 5, 2005
6. Feb 5, 2005

### Muzza

Last edited by a moderator: May 1, 2017
7. Feb 5, 2005

### Townsend

Thanks for the links....I will spend some time studying them. I suppose that I would eventually have reached a stoping point or come to the same conclusion about a being proportional to b. That or made a mistake somewhere and thought I had something I really did not have.

I still want to see where my reasoning takes me.

Regards

Last edited by a moderator: May 1, 2017
8. Feb 5, 2005

### Townsend

I will just continue on from post #5 I made earlier, some family stoped by but they are gone now so I am back to it. By the way I am not sure where I am going with this but I think I can get to where you need to be. Please feel free to just jump in and add to anything I have.

So far I have shown that one way to equality to hold is for $n=2\\\mbox{ when }\\\ {a_1b_2=a_2b_1}$

Well I worked out the gory details to check this for n=3 and there was no surprises. After simplifying, factoring and all that goodness I ended up with

$$0 \leq (a_1b_2-a_2b_1)^2+(a_1b_3-a_3b_1)^2+(a_2b_3-a_3b_2)^2$$

Which leaves me with a hypothesis that one way for equality to hold we need the following conditions to hold.

For each $$a_n$$

$$\frac{a_{n-1}}{a_n} = \frac{b_{n-1}}{b_n}$$

I am going to try to show this is true for all n in the set of natural numbers. I will leave all that out of here but basically I am going to try induction to see where it takes me. If we can show this is true perhaps we can use normalization to get to where we want to go. I am sure this has been done before and it may well be a waste of my time to go through all of this but at the very least I find it fun to see what I can do.

I'll be back, as the govenator would say....

Regards

Last edited: Feb 5, 2005
9. Feb 5, 2005

### MathStudent

Townsend's missing code

Townsend wrote:

$$\frac{a_{n-1}}{a_n} = \frac{b_{n-1}}{b_n}$$

note to townsend: you left out a curly bracket

10. Feb 5, 2005

### Townsend

I got it....in trying to show this is true by induction I came up with some interesting results.

My hypothesis was that for the Cauchy inequality to have equality we must have

$$\frac{a_1}{a_2}=\frac{b_1}{b_2}$$

$$\frac{a_1}{a_3}=\frac{b_1}{b_3}$$

and so on through n. And then the same squence of equations for a_2 and b_2 and then the same through a_3, b_3 until we get to a_{n-1}, b_{n_1} in which we just have

$$\frac{a_{n-1}}{a_n} = \frac{b_{n-1}}{b_n}$$

Well when we add these equations together we end up with a very complicated looking expresion. I will try to write it out so you can see it but then I am going to try to normalize the series to make things look much cleaner. Not too sure if I can push this thing around like I wish but I will do my best.

I am going to break this up into the LHS and the RHS just to make it easier.

$$(\frac{a_1}{a_2}+\frac{a_1}{a_3}+\cdots+\frac{a_1}{a_n})+(\frac{a_2}{a_3}+\frac{a_2}{a_4}+\cdots+\frac{a_2}{a_n})+\cdots+\frac{a_{n-1}}{a_n}$$

Now the LHS looks the same only with b's instead of a's

$$(\frac{b_1}{b_2}+\frac{b_1}{b_3}+\cdots+\frac{b_1}{b_n})+(\frac{b_2}{b_3}+\frac{b_2}{b_4}+\cdots+\frac{b_2}{b_n})+\cdots+\frac{b_{n-1}}{b_n}$$

So in my mind we need to see how to make the LHS and the RHS into a single fraction and at the same time hopefully simplify the equation a bit. This part is kind of hard, at least for me it is, so I am going to leave you with this until I can make some more sense out of everything.

Regards

Last edited: Feb 5, 2005
11. Feb 7, 2005

### Galileo

Without vectors, the usual way this is proven is:

(I): Assume x and y are proportional. It's easy to show equality holds.
(II): Assume they are not proportional: $\lambda y-x\not=0$ for all $\lambda \in \mathbb{R}$.

$$0<|\lambda y-x|^2=\sum_{i=1}^n(\lambda y_i-x_i)^2$$

expand the right side. You'll get a quadratic equation in $\lambda$ which is never 0, hence it doesn't have any real roots and it's discriminant is therefore negative. Write the discriminant down, the solution is immediate.