Prove that if 0 < a < b,
then a < √ab < (a + b)/2 < b
To prove this use the 12 properties of numbers (commutativity, trichotomy law, associativity, etc...).
The Attempt at a Solution
The main problem is I don't know if I need to define the square root function since I'm doing a proof. I'm new to all this, this is my first time doing analysis, so bear with me.
I assume you break up the inequality into parts, since a<b is already defined, we don't need to prove the whole inequality. So I did the first part, unintutively
a^2 < (ab)
(a)(a) (a^-1) < (ab)(a^-1)
a<b which was assumed thus the inequality is proven
(a + b)/2 < b
(a +b)(2^-1) (2) < b(2)
a < b
Problem is I don't know what to write division of 2 as, since it's not listed in basic property. Also I don't know how to prove the 2 centre terms of the inequality: √ab < (a + b)/2