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I need to show $a^2 +b^2 > 2ab$ , but only with natural numbers, for that reason, I can't use negative numbers, the zero, or others non-natural numbers, e.g. I can't use the fact $(a-b)^2 > 0$

uniquesolution
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Jorge S.
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    If you can't use $(a-b)^2 > 0$, can you use $(b-a)^2 > 0$ instead? – Sebastian Schulz Mar 26 '21 at 21:52
  • Use AM-GM then. – Anon Mar 26 '21 at 21:54
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    Usually inequality axioms involve comparisons with 0. What kind of axioms do you suggest to use? – user Mar 26 '21 at 21:54
  • @Kaind And how would you prove AM-GM? I would start with this inequality, then work from there. So to me, using AM-GM here would be circular. – Arthur Mar 26 '21 at 21:54
  • @Arthur Proof using induction on $n$ (no. of variables). – Anon Mar 26 '21 at 21:55
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    Also, obviously the inequality is false for $a=b$ (even if they aren't $0$). But then again, no zero allowed. –  Mar 26 '21 at 21:55
  • The question is ill posed. It is not true, for example, if $a=b$. – uniquesolution Mar 26 '21 at 21:57
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    @Kaind And what's the base case? To me, it's this exact inequality (or maybe a single variable if you're feeling trivial, but I like starting at 2) . What's the induction step? To me, it's this exact inequality. So that makes it circular. – Arthur Mar 26 '21 at 21:59

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One of $a$ or $b$ is larger, so say $a<b.$ Then for some $c$, $b=a+c.$ Substitute this in to the left side:

$$a^2+b^2 = a^2 +(a+c)^2 = 2a^2+2ac+c^2 > 2a^2+2ac = 2a(a+c) = 2ab.$$

B. Goddard
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  • Doesn't the existence of $c$ require $0$? I think the OP must mention which framework they are working with - clearly it is something more primitive that Peano's axioms. – Anon Mar 26 '21 at 22:03
  • Is the relation $a<b$ defined as an existence of $c$? – user Mar 26 '21 at 22:05
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    @Kaind He said "natural numbers". So I'm assuming Peano but just not allowed to use zero. $b$ is some successor of some successor of ..... $a$. $c$ is how many times you write "successor." – B. Goddard Mar 26 '21 at 22:13