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I am a non-mathematician attempting to read and work through Spivak's Calculus by myself. So for the problem in the subject title, Spivak employs a proof by contradiction method in his solutions (and some have answered it directly, by factoring or completing the square). But when I attempted it, I came up with a direct proof as below (pardon the informal language and my English):

We need to prove $4x^2 + 4y^2 + 6xy > 0, \ x \neq 0 \neq y$. We know that $4x^2, 4y^2 \in P$, where P is the positive set. If $6xy \in P $ or $6xy = 0$, our expression is positive. However if $6xy \notin P$, then we need to prove $|6xy| < 4x^2 + 4y^2$

We already know that $ 4x^2+4y^2+8xy = (2x+2y)^2 > 0$ $ \implies |8xy| < 4x^2 + 4y^2 \space \forall \space 8xy $

Since $|8xy| < 4x^2 + 4y^2 $, and since $ |6xy| < |8xy| $, we have $|6xy| < 4x^2 + 4y^2$. Therefore, $4x^2 + 4y^2 + 6xy > 0 \space \forall \space 6xy $

So, please tell me if this is a correct attempt at proof. If it is, why does Spivak use a proof by contradiction method?

tychonovs-scholar
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Vignesh M
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  • Be careful with your $>$ versus $\ge$. $4 x^2 + 4 y^2 + 6 x y = 0$ when $x = y = 0$. So $4 x^2 + 4 y^2 + 6 x y > 0$ is not necessarily true. – Robert Israel Jan 17 '25 at 16:07
  • Ah yes. Thank you for the suggestion. Made the edit. – Vignesh M Jan 17 '25 at 16:14
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    Note that rather simpler is to rewrite as $3(x+y)^2+x^2+y^2$ – Mark Bennet Jan 17 '25 at 16:44
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    Mark Bennet illustrates a good general rule. If you want to prove a minimum, say all values are above 0, it's helpful to break things up in to sums of squared expressions as much as possible. – TurlocTheRed Jan 17 '25 at 19:28
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    @TurlocTheRed The situation for writing a non-negative polynomial as a sum of squares is the subject of Hilbert's 17th problem (see https://en.wikipedia.org/wiki/Hilbert%27s_seventeenth_problem). In this case - homogeneous in two variables (equivalent to a polynomial in a single variable) a sum of squares of polynomials can always be found. In more complicated cases squares of rational functions are required (Artin). It is useful to know that it can always be done. – Mark Bennet Jan 17 '25 at 22:14

1 Answers1

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This is not yet quite correct:

  • It is not true that $(2x+2y)^2 > 0$, even if $x\neq 0 \neq y$. Take $x = 1$ and $y = -1$, for example.
  • The exercise actually asks something a little tiny bit stronger: we need to prove that $4x^2 + 6xy + y^2 > 0$ if either $x\neq 0$ or $y\neq 0$.
  • Minor style issue: it is best not to use symbols like $\implies$ and $\forall$ in your proofs, unless (a) you are being informal or (b) they are part of a formula you are manipulating. Expressions like $\forall 8xy$ are sloppy (you should say something like "for all real numbers $x,y$"). The arrow $\implies$ is also potentially confusing: I can say "We know that $0 = 1 \implies 4x^2 + 6xy + y^2 > 0$", and I wouldn't be lying, since an incorrect statement implies any other (look up "ex falso quodlibet" or "principle of explosion").

The exercise can be proved using both direct reasoning and proof by contradiction. Spivak almost certainly chose contradiction because he thought the proof would be more concise.

Mentastin
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  • Agreed. The expression that needs to be proved is 4x^2+4y^2+6xy > = 0 indeed. I should have been more careful while writing that. So, if I understand you correct, what you are saying is that we can still prove the expression by applying the same logic that I used here, but my proof lacks the brevity that Spivak's solution has? Is that correct? Edit: Noted your edit on the style issue. Point taken. Will remember that as I proceed to do more proof-based maths. – Vignesh M Jan 17 '25 at 16:53
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    @VigneshM Spivak's exercise was to prove $4x^2 + 4y^2 + 6xy > 0$. (Not $\geq 0$!) It is possible to fix your proof (for example you could separate out the case $x = -y$ and deal with it separately). The result will probably be a proof that is longer than Spivak's. – Mentastin Jan 17 '25 at 16:59
  • Just checked again. Yes. You are correct. I replied without referring back to the actual question from the book. Now I get what you mean. Thank you. – Vignesh M Jan 17 '25 at 17:02