Show that if $a,b \in \Bbb R$ then $\max\{a,b\} = \frac12(a+b+|a-b|)$ and $\min\{a,b\} = \frac12(a+b-|a-b|)$

how would you go about solving this?

I started with suppose $a \leq b$

Also, show min{a,b,c} = min{min{a,b},c}.

How would I go about showing that?
Suppose  that $a\le b$.

$$\frac{a+b+|a-b|}{2}=\frac{a+b+b-a}{2}=b=\max \{a,b\}$$

And 

$$\frac{a+b-|a-b|}{2}=\frac{a+b-(b-a)}{2}=a=\min \{a,b\}$$
You can use the fact that $$\min\{x,y\}=-\max\{-x,-y\}$$ to get the result about mins from the one about maxes.