Without loss of generality suppose a>=b, then

(1/2)[a+b+abs(a-b)]=(1/2)[a+b+a-b]=a.

If b were the larger we would observe that max(a,b)=max(b,a) then

we would have:

max (b,a) = (1/2)[a+b+abs(b-a)],

and the result would again follow.

This is similar.min (a,b) = (1/2)[a+b-abs(a-b)]

RonL