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I was thinking that it is possible to answer this Math.SE question here using the trichotomy of the real numbers but I got some logic trouble.

The question was. Prove that for all $x,y > 0$ integers is true that $$\sqrt{xy} \leq \frac{x+y}{2}$$

Then I think in this proof.

We can use the Trichotomy of the real numbers . Given the two numbers $\sqrt{xy}, \frac{x+y}{2} \in \mathbb{R}$ and the usual order relation in $\mathbb{R}$, there are just three possibilities: $$\sqrt{xy}> \frac{x+y}{2}\tag{1}$$ $$\sqrt{xy} = \frac{x+y}{2}\tag{2}$$ $$\sqrt{xy}< \frac{x+y}{2}\tag{3}$$

Let $x,y > 0$ be integers. Suppose that for all $x,y$ we have $$\sqrt{xy}>\frac{x+y}{2}$$

then we get that, squaring both sides,

$$xy > \frac{1}{4}(x^2+2xy+y^2) \implies 4xy > x^2+2xy + y^2 \implies 2xy > x^2+y^2$$

This is absurd, if $x = y$ we have $2x^2 > 2x^2$. So we conclude that for general $x,y \geq 0$ we have that

$$\sqrt{xy} \leq \frac{x+y}{2}$$

Question: Does that proves the first question? Seems that it only proves that the relation '$>$' is not possible. But it does not prove that '$\leq$' is true for all $x,y>0$ integers. It could be possible to proof using this method but in other way?

R.W
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  • Are you having difficulty with the notion that when $A \gt B$ is impossible for real numbers $A,B$, then it must be true that $A \le B$? You say so much about trichotomy that I'm finding it difficult to see what troubles you about this. – hardmath Oct 08 '17 at 21:48
  • @hardmath I thought my question was clear, is my proof and way of thought correct? What you've just sayed is clear to me but when we are dealing with all numbers that satisfy a relation this can also be used? – R.W Oct 08 '17 at 21:51
  • It troubles me that doesn't seems to me that I've just proved what the question asked to prove. Seems that it can be possible that some $x,y> 0$ doesn't satisfy $\leq$ property. – R.W Oct 08 '17 at 21:52

2 Answers2

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Your reasoning seems to be wrong, because you want to prove that $$\sqrt(xy) \leq (x+y)/2,$$ for all $x,y$ and you try to do it by supposing that $\sqrt(xy) > (x+y)/2$ for all $x, y$ and get a contradiction.
This is a major logical flaw, because the negation of $$\forall x \Phi(x)$$ is not $\forall x \neg \Phi(x)$ but rather $$\exists x \neg \Phi(x).$$ So in this case, if you insist in following this proof by contradiction, what you need is to get a contradiction from the hypothesis that for some $x_0, y_0$, $$\sqrt(x_0y_0) > (x_0+y_0)/2,$$ which, of course, should not be difficult, but it seems that a direct proof is just as easy and more elegant, in this case...

amrsa
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  • Then, when showing that the $>$ inequality is not possible for some $x_0,y_0$ then I can conclude that for all $x,y$ the $\leq$ inequality is true? Thank's this was exactly what I was looking for. – R.W Oct 09 '17 at 18:26
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    Yes, that's it. But the important thing is to realise the logical error that you made, which I trust you don't make often. It happens... :) – amrsa Oct 09 '17 at 18:38
  • Totally. I have to negate things correctly Thanks very much! – R.W Oct 09 '17 at 18:42
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I think you just have your order of things interchanged. Your proof as written supposes the hypothesis "for all $x,y$, we have $\sqrt{xy} > (x+y)/2$".

Instead, what you want is, for every $x,y$, to suppose the hypothesis $\sqrt{xy} > (x+y)/2$

More schematically, the incorrect argument has the form

  • Suppose $\forall x,y > 0: \sqrt{xy} > (x+y)/2$
    • ...
    • Therefore, contradiction
  • Conclude $\neg \forall x,y > 0: \sqrt{xy} > (x+y)/2$

The correct argument I think you were trying to make is

  • Let $x,y$ be positive reals
    • Suppose $\sqrt{xy} > (x+y)/2$
      • ...
      • Therefore, contradiction
    • Conclude $\sqrt{xy} \leq (x+y)/2$
  • Conclude $\forall x,y > 0: \sqrt{xy} \leq (x+y)/2$