# Spivak's Calculus 1-19 Schwarz Inequality

## Problem

There's a question in the third edition of Spivak's Calculus on the Schwarz inequality. It is presented as: $$x_1y_1+x_2y_2 \leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$$ Part (a) asks: (i) to prove that if $$x_1=\lambda y_1$$ and $$x_2=\lambda y_2$$ then equality holds, (ii) proving equality for $$y_1=y_2=0$$, and (iii) to prove the inequality given $$0 < (\lambda y_1-x_1)^2 + (\lambda y_2-x_2)^2 = \lambda^2(y_1^2+y_2^2) - 2\lambda(x_1y_1 + x_2y_2) + (x_1^2 + x_2^2) \tag{*}\label{*}$$ and using Problem 18 (which relates to proving the quadratic formula, $$b^2-4c$$, minimum of $$ax^2 + bx + c$$, etc.)

## Part (a)(i)

I went about part (i) by multiplying out the RHS and then squaring both sides to give $$\lambda^2y_1^4+2\lambda^2y_1^2y_2^2+\lambda^2y_2^4=\lambda^2y_1^4+2\lambda^2y_1^2y_2^2+\lambda^2y_2^4$$

## Part (a)(ii)

For part (ii) I substituted $$0$$ for $$y_1$$ and $$y_2$$ and ended with $$0=0$$ by way of $$0+0=\sqrt{x_1^2+x_2^2}\sqrt{0^2+0^2}$$ although something didn't feel right about this so I'm not sure if it is correct.

## Part (a)(iii)

Part (iii) I struggled with and is the main reason I am asking this question (I understand it has been asked many times before but trying to solve it has made me realize there are quite a few things I don't understand).

I noticed that \ref{*} is equivalent to $$ax^2 + bx + c > 0$$ so (because of Problem 18) went about trying to show that $$b^2-4c<0$$ but it became clear that this wasn't true. I assume it's because this only works when $$a=1$$?

By completing the square I got $$(y_1^2+y_2^2)^2(\lambda+\frac{x_1y_1+x_2y_2}{y_1^2+y_2^2})^2 + x_1^2+x_2^2-\frac{x_1y_1+x_2y_2}{2}>0$$ I then decided to deal with it on a case by case basis of finding the minimum. First, where $$y_1^2+y_2^2=0$$, giving zero in the first set of brackets. This case was covered earlier but gives $$x_1^2+x_2^2-\frac{x_1(0)+x_2(0)}{2}>0$$. This seems to contradict my earlier result of $$0=0$$ as any case where $$x_1 \ne 0$$ or $$x_2 \ne 0$$ would give an answer $$> 0$$.

For the second case, where the second set of brackets equals zero, I came up with the condition that $$\frac{x_1y_1+x_2y_2}{y_1^2+y_2^2}$$ = $$-\lambda$$ and so $$c=\frac{\lambda}{2}(y_1^2+y_2^2)$$, proving the inequality so long as $$\lambda > 0$$.

## Part (b)

This asks for a proof of the Schwarz inequality using $$2xy \le x^2 + y^2$$ with $$x=\frac{x_i}{\sqrt{x_1^2+x_2^2}}, y=\frac{y_i}{\sqrt{y_1^2+y_2^2}}$$ for $$i =1$$ and $$i = 2$$.

I think I am being very naive about this but my instinct wants to say that since $$2xy \le x^2 + y^2$$ is derived from $$(x - y)^2 \ge 0$$ it must automatically be true, but part of me thinks this would only work because $$x$$ and $$y$$ have been defined as given. Maybe someone could shed a little more light on this?

## Questions

I will try to highlight here some key questions.

1. Are parts (a)(i)-(iii) correct?
2. Am I right in saying that $$ax^2 + bx + c > 0$$ for all $$x$$ when $$b^2 - 4c < 0$$ iff $$a=1$$?
3. Is it true that (a)(iii) should only work when $$\lambda > 0$$ or have I made a mistake?
4. My naive assumption (and possible reason for why it isn't) in part (b) for the Schwarz inequality being "automatically" true because of $$(x-y)^2 \ge 0$$.

Finally, thank you very much for reading this and any help (especially in terms of my thinking) would be greatly appreciated. I am new to this, but enjoying it, so the techniques and ways of thinking are taking some time to set in but hopefully good practice will solve that!

• Specific your inequality we can prove much more easier. – Michael Rozenberg Jul 20 '19 at 21:23
• @MichaelRozenberg Sorry, I don't quite understand. The Schwarz inequality is listed at the top of the post, but I am mostly asking about a number of problems I came across in trying to prove it through a few textbook questions that are a result of my misunderstandings. Thanks for your reply and I hope my question is clearer now. – Rory Jul 20 '19 at 23:10

For starters, consider factoring in the first part of (a). The second part of (a) also seems right. In a set of problems such as this, something that seems fairly straightforward will usually be very helpful in some surprising manner in later non-trivial parts of the problem. Now, for Part iii you did not mention that there is no number 𝜆 such that 𝜆$$x_1$$ = $$y_1$$ and 𝜆$$x_2$$ = $$y_2$$, which is important (otherwise, it wouldn't have been mentioned). It should be somewhat easier to go from here but keep trying.