Proving $a \leq \frac{a+b}{2} \leq b$ I want to prove the following $a \leq \frac{a+b}{2} \leq b$, where we know that $0 \leq a \leq b$. My proof goes as follows.
Suppose $a \leq \frac{a+b}{2} \leq b$, then we know $a \leq \frac{a+b}{2}$ and this implies that $\frac{a}{2} \leq \frac{b}{2}$ and so, since this is true because $ a \leq b$, then it is true that $a \leq \frac{a+b}{2} \leq b$. 
Would this be a right approach to do the proof?
Thanks!
 A: You can't suppose $a \leq \frac{a+b}{2} \leq b$ because this what you want to prove.
First proof
First assume that $a\leq b$, then try to prove your inequality (subtract $\frac{a+b}{2}$ from what you want  and see what happens).
Second proof
By definition, $$[a,b]=\{at+(1-t)b\mid t\in [0,1]\}.$$
See what happen for $t=\frac{1}{2}$ and conclude.
A: Hint: $2a = a + a \leq a + b \leq b +b = 2b$
A: Here's another approach:
$$a=\frac{a+a}2\le\frac{a+b}2\le\frac{b+b}2=b.$$
A: Yet another approach: start with what you want to prove, and simplify, then relate it to the assumptions you're allowed to use.$
\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}}
$
In other words, calculate as follows:
$$\calc
    a \le \tfrac{a+b}{2} \le b
\op=\hints{notation: split into two inequalities;}\hint{arithmetic: multiply both inequalities by 2}
    2a \le a+b \;\land\; a+b \le 2b
\op=\hint{arithmetic, simplify: subtract $\;a\;$ in LHS, and $\;b\;$ in RHS}
    a \le b \;\land\; a \le b
\op=\hint{logic: simplify}
    a \le b
\endcalc$$
So you were trying to show $$
0 \le a \le b \;\then\; a \le \tfrac{a+b}{2} \le b
$$ but the above proof shows that the stronger  $$
a \le b \;\equiv\; a \le \tfrac{a+b}{2} \le b
$$ holds.
