# More Schwarz inequality proofery

• swevener
In summary: So the inequality is because x_{1} y_{1} + x_{2} y_{2} \geq (x_{1} y_{1} + x_{2} y_{2})^{2}. Basically, the more complex the variables, the greater the inequality.
swevener

## Homework Statement

Prove the Schwarz inequality by first proving that

$$(x_{1}^{2} + x_{2}^{2})(y_{1}^{2} + y_{2}^{2}) = (x_{1} y_{1} + x_{2} y_{2})^{2} + (x_{1} y_{2} - x_{2} y_{1})^{2}.$$

## Homework Equations

$$x_{1} y_{1} + x_{2} y_{2} \leq \sqrt{x_{1}^{2} + x_{2}^{2}} \sqrt{y_{1}^{2} + y_{2}^{2}}$$

## The Attempt at a Solution

I'm not sure if my logic is right. I did the little proof above, and with that I can say

$$-(x_{1} y_{2} - x_{2} y_{1})^{2} \leq (x_{1} y_{1} + x_{2} y_{2})^{2}.$$
Can I then sweep the LHS under the zero and say

$$(x_{1} y_{1} + x_{2} y_{2})^{2} \leq (x_{1}^{2} + x_{2}^{2})(y_{1}^{2} + y_{2}^{2}),$$
then take the square root to finish the proof?

You don't need your relevant equation!

You can prove the equality holds by simply starting with one side of the equation (remember not to assume they are equal, that would be assuming what you're trying to prove!) and end up with the other.

You can start with either side to get the other but I'd suggest starting with the right hand side.

See what you come up with!

I did that part. I'm asking if I got from there to the Schwarz inequality, the "relevant equation", logically.

Okay, sorry, I didn't see that line of text.

So what you're doing (I think you left this step out) is saying that $(x_{1} y_{1} + x_{2} y_{2})^{2} + (x_{1} y_{2} - x_{2} y_{1})^{2} \geq 0$ because the values are both positive and then you subtracted the one term. I'm still not sure how you went from the third equation in your post to the final equation. I think you can word that a little bit better? I get the sense you have to the right idea and understand why it's true, it just isn't obvious to me from reading it why you did it.

Yeah, I started with

$$0 \leq (x_{1} y_{1} + x_{2} y_{2})^{2} + (x_{1} y_{2} - x_{2} y_{1})^{2},$$
and since being squared mean both terms will always be positive, multiplying one of them by -1 makes that one always less than or equal to the other:

$$-(x_{1} y_{2} - x_{2} y_{1})^{2} \leq (x_{1} y_{1} + x_{2} y_{2})^{2}.$$
Although looking at it now, I think I threw that in there just so I didn't disappear the LHS without comment. What I had in mind was this sort of thing:

$$-(x_{1} y_{2} - x_{2} y_{1})^{2} \leq (x_{1} y_{1} + x_{2} y_{2})^{2} \leq (x_{1}^{2} + x_{2}^{2})(y_{1}^{2} + y_{2}^{2})$$
And then just cut out the $-(x_{1} y_{2} - x_{2} y_{1})^{2}$, because I'm not sure how else to get rid of it.

Okay, I get what you are saying now. I think this way might be a little easier, you can be the judge.

If $a^{2} = b^{2} + c^{2}$, then $a^{2} \geq b^{2}$. I don't see any reason to "justify" getting rid of the $c^{2}$ term, it's just extra that you don't need.

So, this is basically what you are doing except not trying to fiddle with the $c^{2}$ term. All you have to do is explain why an inequality appears when you get rid of the $c^{2}$ term.

Okay, thanks.

## 1. What is the Schwarz inequality?

The Schwarz inequality, also known as the Cauchy-Schwarz inequality, is a mathematical inequality that relates the inner product of two vectors to their lengths. It states that the absolute value of the inner product of two vectors is less than or equal to the product of their lengths.

## 2. What is the significance of the Schwarz inequality?

The Schwarz inequality is significant because it has numerous applications in various fields of mathematics, including linear algebra, functional analysis, and probability theory. It is also a fundamental tool in proving other important mathematical theorems.

## 3. How is the Schwarz inequality proved?

The Schwarz inequality can be proved using various methods, including geometric, algebraic, and functional analysis approaches. The most common method is by using the Cauchy-Schwarz inequality in the context of complex numbers and then generalizing it to real vectors.

## 4. Can the Schwarz inequality be extended to more than two vectors?

Yes, the Schwarz inequality can be extended to any finite number of vectors. This is known as the generalized Schwarz inequality or the n-dimensional Schwarz inequality. It is often used in multivariable calculus and optimization problems.

## 5. Is there a reverse version of the Schwarz inequality?

Yes, there is a reverse version of the Schwarz inequality, known as the reverse Schwarz inequality or the reverse Cauchy-Schwarz inequality. It states that the absolute value of the inner product of two vectors is greater than or equal to the product of their lengths. This inequality is useful in proving the existence of orthogonal vectors.

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