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!
On
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.
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.