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The algebraic proof is straightforward where we conclude that square of every real number is non-negative, but I can't see anything in the axioms to support that claim. Here is the idea I've had so far: proceed with algebraic manipulations (which follow directly from axioms) to arrive at $0 \leq (\frac{1}{2}x - \frac{1}{2}y)^2=(\frac{1}{2}x - \frac{1}{2}y) \cdot (\frac{1}{2}x - \frac{1}{2}y)$, then proceed with cases: $\begin{cases} i) \quad x \leq y \\ ii) \quad y \leq x \end{cases}$. For $i$) we have $\frac{1}{2}x - \frac{1}{2}y \leq 0 \space(*)$, here I had an idea to produce a $-z \cdot -z$ expression, where $z$ is $(*)$ and then use a fact we proved previously: $(-1)\cdot(-1)=1$, but I'm struggling to come up with a precise continuation. For $ii)$ I'm omitting the reasoning since it's pretty straightforward I believe.

Edit: $\mathbb R$ is a totally ordered field under relation $\leq$ + axiom of completeness, so hopefully this clarifies which axioms are provided.

powerline
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  • But which axioms? – Alma Arjuna Dec 16 '23 at 22:25
  • @AlmaArjuna Sorry for not mentioning. Axioms of the real numbers-i.e. $\mathbb R$ is a totally ordered field under relation $\leq$ + axiom of completeness. – powerline Dec 16 '23 at 22:28
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    Your reasoning can be used to prove the stronger fact: the square of every real number is positive.

    In fact, if $z\ge0$, then $z^2\ge0$. Otherwise, $-z\ge0$ and $z^2=(-1)^2z^2=(-z)^2\ge 0$.

     – Alma Arjuna Dec 16 '23 at 22:40
  • @AlmaArjuna Thanks! I was missing that RHS follows from LHS in $z \leq 0 \leq -z$ for $z=\frac{1}{2}x - \frac{1}{2}y$ – powerline Dec 16 '23 at 22:50
  • I don’t follow or understand how the $\frac12$’s help. Why not aim for proving $4xy\le (x+y)^2$? You should not need any case analysis whatsoever. – Ted Shifrin Dec 16 '23 at 23:07
  • @TedShifrin It wasn't my intention to mean that the $\frac {1}{2}$'s helped, my brain forgets about a lot of algebra when I'm supposed to use only axioms - I just missed what you're saying - it's definitely cleaner and better, thanks! – powerline Dec 16 '23 at 23:31
  • @TedShifrin I used the cases because they helped me to flesh out the idea I had, please do leave a hint for an approach without cases. thanks – powerline Dec 16 '23 at 23:36
  • As others have said, it is just a consequence of the fact that the square of any real number is nonnegative. For positive numbers, it is closure under multiplication. For negative numbers, you need $(-1)(-1) = -(-1)=1$. So the lemma is that $(-1)a= -a$. How do you prove that? – Ted Shifrin Dec 16 '23 at 23:40
  • $(-1)a= (-1)a + a - a = ((-1) + 1)a - a = 0a - a = -a $, that's what I used when finishing the case $i)$. I guess the cases aren't needed if we previously established that squares of real numbers are nonnegative but since we didn't do that it made sense to include cases? Sorry if I'm missing something obvious – powerline Dec 16 '23 at 23:51
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    Well, I’m using cases to prove the nonnegativity of the square. You have to there. But that’s it. – Ted Shifrin Dec 17 '23 at 00:31

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$\frac14 (x+y)^2 = \frac14 (x^2 + 2xy + y^2)$ so $xy \leq \frac14 (x+y)^2$ if and only if $0 \leq \frac14 (x^2 -2xy + y^2)$ however we also have that $\frac14 (x^2 -2xy + y^2) = (\frac{x-y}2)^2$.

So all that is left is to show that squares are non-negative. The square of $0$ is $0$ and so non-negative. The axioms state if $a,b >0$ then $ab > 0$ and by setting $b=a$ we have a square so $a^2 > 0$ when $a > 0$. Finally, if $a < 0$ then $-a > 0$ and setting $b=-a$ we have $0 <(-a)^2 = (-1)^2a^2 = a^2$ and so in all three cases squares are non-negative and we are done.

Note that the fact $x,y \geq 0$ is unnecessary and it holds for all reals.

CyclotomicField
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  • "and since squares are non-negative" is the critical part where I struggled - because of what axioms? No clue why $x, y \geq 0$ was included in the exercise. Thanks for responding – powerline Dec 16 '23 at 23:59
  • I agree.The nonnegativity assumption on $x$ and $y$ is a red herring. – Ted Shifrin Dec 17 '23 at 00:32
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    @powerline It's straightforward to prove that casewise from the axioms defining $\le$. See the answers on this page. – Jam Dec 17 '23 at 00:44
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    Re. the seemingly superfluous condition that $x$ and $y$ are nonnegative, that is likely there because most proofs that squares are nonnegative proceed by cases for each sign. Eliminating the need for cases makes it a potentially easier proof, albeit more limited. – Jam Dec 17 '23 at 00:47
  • @Jam thanks for the link, the chosen answer was very nice! Though I see now that I wasn't given the $0 < a \land 0 < b \implies 0 < ab$, which is kind of weird I guess – powerline Dec 17 '23 at 01:23
  • @powerline I've edited to include a proof that squares are always positive using the ordered field axioms. – CyclotomicField Dec 17 '23 at 01:35
  • what dioes the 2nd para. even mean? – LÜHECCHEgon Dec 17 '23 at 02:06
  • @JiaoCtagon in the second paragraph I'm proving that for all real numbers $x$ that $x^2 \geq 0$. It's done by cases starting with the positive case, then $0$, then the negative case. – CyclotomicField Dec 17 '23 at 02:59