Intro to Real Analysis I am having trouble proving the following:

if $a < b$, then $a < {a+b\over2} < b$. 

I started with the Trichotomy Property and getting to where $a^2>0$, but then I do not know where to go from there. 
Any suggestions?
 A: HINT: It holds that $\frac{a + a}{2} = a$. If $a<b$, what do you know about $\frac{a +a}{2} + \frac{b-a}{2}$? Can you make a similar argument for the second inequality?
A: Start with $a<b$, add $a$ to both sides then time both sides by $\frac{1}2$. You will get $a< \frac{a+b}2$.
A: Hint: $2a=a+a<a+b<b+b=2b$.
A: In particular, we have the following inequality chain:
$$\min(a,b) \le \frac{2}{\frac{1}a+\frac{1}b} \le \sqrt{ab} \le \frac{a+b}2 \le \sqrt{\frac{a^2+b^2}2} \le \max(a,b).$$
A: $a<b \iff \frac{a}{2}<\frac{b}{2} \iff a-\frac{a}{2}<\frac{b}{2} \iff a<\frac{a}{2}+\frac{b}{2}=\frac{a+b}{2}$
Similary $\frac{a+b}{2}<b$.
A: Seeing as $(a+b)/2$ is the avg of $a$ & $b$...
The $a \lt b$ gives you a nice ordering, telling you $a$ is closest to $-\infty$. Thereby, the avg of $a$ & $b$ will be between $a$ & $b$ as opposed to being between $b$ & $a$.
A: Given $a<b$,
$$a<b \Longleftrightarrow 2a<a+b \Longleftrightarrow a<\frac{a+b}{2}$$
$$a<b \Longleftrightarrow a+b<2b \Longleftrightarrow \frac{a+b}{2} < b$$
$$a<\frac{a+b}{2}<b$$
as desired.
