In proving these inequalities, what I did was to take each of the "<" relations and prove them. Is this a valid way of proving if we have got several inequalities as in this problem?
So here is my proof by parts.
$a<\sqrt{ab}$:
Let $P$ be the set of all positive numbers. Since $a,b,b-a\in P$, we have $a(b-a)=ab-a^2\in P$, so $a^2<ab$, which implies that $a<\sqrt{ab}$. (I make a separate proof for this implication, which goes as follows: $a^2<b^2 \iff 0<b^2-a^2=(b-a)(b+a)\in P$, so $(b-a)\in P$, thus $a<b$). So this proves $a<\sqrt{ab}$.
$\displaystyle\sqrt{ab}<\frac{a+b}{2}:$
We have $\displaystyle\sqrt{ab}<\frac{a+b}{2}\iff2\sqrt{ab}<a+b\iff0<a-2\sqrt{ab}+b=(\sqrt{a}-\sqrt{b})^2$. Since $a,b$ are both non-zero, so $0<(\sqrt{a}-\sqrt{b})^2$ holds, thus, proving $\displaystyle\sqrt{ab}<\frac{a+b}{2}$.
$\displaystyle\frac{a+b}{2}<b$:
By closure under addition, $a<b \implies a+b < b+b = 2b$, dividing by $2$, $\displaystyle\frac{a+b}{2}<b$.
Does this work? If not, where did I go wrong? Is there a much better way to show this?
