Suppose that 0 < a < b. Prove that $a < \sqrt{ab} < b $ and $\sqrt{ab} \leq \frac{1}{2}(a+b)$ For part 1, I have used the NOT operator on it, giving me $a \geq \sqrt{ab} \geq b$, and then tried to prove a contradiction to the assumption. I came up with $a = b$ by transitivity, which contradicts our assumption. I'm not sure if this is right. 
For part 2, I have tried proving it directly through algebra, and it didn't work, so I'm fairly certain a different proof method must be used: either contrapositive or contradiction. But that's basically as far as I can get. I can't seem to make anything logical happen via algebra.
 A: 
For the second part, this expression is equivalent to $4ab \leq a^2 + 2ab + b^2$ since all values are positive, and this is equivalent to $0 \leq a^2 -2ab + b^2 = (a-b)^2$ which is again equivalent to $0 \leq b-a$ since $b>a$ and is equivalent to $a\leq b$.  
So apparently this should be a strict inequality.

Edit:  This has some circular logic at the end.  To correct it, instead do this:  
Suppose $0<a<b$. Then you get $0<b−a$ so $0<(b−a)^2=(a−b)^2=a^2−2ab+b^2$ which implies $4ab<a^2+2ab+b^2=(a+b)^2$ which implies $2\sqrt{ab}<a+b$ which implies $\sqrt{ab}<(a+b)/2$. 
A: We seek to prove $a<\sqrt{ab}<\frac{a+b}{2}<b$
We take the inequalities one at a time
Firstly, since $a<b, \sqrt{ab}>\sqrt{aa}=a$  
Note that when expanding $(\sqrt{a}-\sqrt
{b})^2$, which is clearly greater than zero, we obtain $a+b-2\sqrt
{a}\sqrt{b}>0$, thus $a+b>2\sqrt{a}\sqrt{b}$ and $\frac{a+b}{2}>\sqrt{ab}$
Finally, since $a<b, \sqrt{ab}<\sqrt{bb}=b$
Combining these, we obtain the desired inequalities  
A: I would start with what you need to prove, and try to simplify it using the information you have.$%
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
%$
For part 1, we could calculate like this:
$$\calc
    \tag{1}
    a < \sqrt{ab} < b
\op\equiv\hints{square both sides of both inequalities,}\hints{no sign changes since all are positive}\hint{-- to get rid of the square root}
    a^2 < ab < b^2
\op\equiv\hints{divide left inequality by positive $\;a\;$;}\hints{divide right inequality by positive $\;b\;$;}\hint{-- this is the most direct simplification}
    a < b \;\land\; a < b
\op\equiv\hint{simplify}
    a < b
\endcalc$$
So the complex $\Ref{1}$ turns out to be equivalent to part of the assumption.
And for part 2:
$$\calc
    \tag{2}
    \sqrt{ab} \leq \frac{1}{2}(a+b)
\op\equiv\hints{double then square both sides of the inequality,}\hints{no sign change since both are positive}\hints{since $\;a\;$ and $\;b\;$ are positive}\hint{-- to get rid of the fraction and square root}
    4ab \leq a^2+2ab+b^2
\op\equiv\hint{simplify}
    a^2-2ab+b^2 \geq 0
\op\equiv\hint{simplify -- the simplest thing to do with this}
    (a-b)^2 \geq 0
\op\equiv\hint{always $\;x^2 \geq 0\;$}
    \true
\endcalc$$
And note that for this second part we did not need the assumption that $\;a<b\;$.
