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?
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...