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?


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!

  • $\begingroup$ Specific your inequality we can prove much more easier. $\endgroup$ – Michael Rozenberg Jul 20 '19 at 21:23
  • $\begingroup$ @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. $\endgroup$ – 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.


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