How can I prove $a^2 +b^2 >2ab$ only with Natural numbers?

Question

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 $(a-b)^2 > 0$

Answer 1

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:

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