Nice expression for minimum of three variables? As we saw here, the minimum of two quantities can be written using elementary functions and the absolute value function.
$\min(a,b)=\frac{a+b}{2} - \frac{|a-b|}{2}$
There's even a nice intuitive explanation to go along with this: If we go to the point half way between two numbers, then going down by half their difference will take us to the smaller one. So my question is: "Is there a similar formula for three numbers?"
Obviously $\min(a,\min(b,c))$ will work, but this gives us the expression:
$$\frac{a+\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)}{2} - \frac{\left|a-\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)\right|}{2},$$
which isn't intuitively the minimum of three numbers, and isn't even symmetrical in the variables, even though its output is. Is there some nicer way of expressing this function?

First, define
$$
\Delta=|a-b|+|b-c|+|c-a|\newcommand{\Mu}{\mathrm{M}}\tag{1}
$$
It is somewhat intuitive that
$$
\frac{\Delta}{2}=\max(a,b,c)-\min(a,b,c)\tag{2}
$$
For example, if $a\ge b\ge c$ then $|a-b|+|b-c|+|c-a|=2a-2c$.
Next, define
$$
\Sigma=a\left(1-\frac{|b-c|}{\Delta}\right)+b\left(1-\frac{|c-a|}{\Delta}\right)+c\left(1-\frac{|a-b|}{\Delta}\right)\tag{3}
$$
Again, if $a\ge b\ge c$, then
$$
\begin{align}
\Sigma
&=a\left(\frac{2a-b-c}{2(a-c)}\right)+b\left(\frac{a-c}{2(a-c)}\right)+c\left(\frac{a+b-2c}{2(a-c)}\right)\\
&=a+c
\end{align}
$$
Thus, considering the symmetry of $(3)$, it is evident that
$$
\Sigma=\max(a,b,c)+\min(a,b,c)\tag{4}
$$
Combining $(2)$ and $(4)$ yields
$$
\max(a,b,c)=\frac{\Sigma}{2}+\frac{\Delta}{4}\tag{5}
$$
and
$$
\min(a,b,c)=\frac{\Sigma}{2}-\frac{\Delta}{4}\tag{6}
$$
At least $(5)$ and $(6)$ are symmetric in $a$, $b$, and $c$ since $(1)$ and $(3)$ are. That is,
$$
\begin{align}
\max(a,b,c)
&=\frac{a}{2}\left(\frac{|c-a|+|a-b|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{b}{2}\left(\frac{|a-b|+|b-c|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{c}{2}\left(\frac{|b-c|+|c-a|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{|a-b|+|b-c|+|c-a|}{4}\tag{7}
\end{align}
$$
and
$$
\begin{align}
\min(a,b,c)
&=\frac{a}{2}\left(\frac{|c-a|+|a-b|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{b}{2}\left(\frac{|a-b|+|b-c|}{|a-b|+|b-c|+|c-a|}\right)\\
&+\frac{c}{2}\left(\frac{|b-c|+|c-a|}{|a-b|+|b-c|+|c-a|}\right)\\
&-\frac{|a-b|+|b-c|+|c-a|}{4}\tag{8}
\end{align}
$$
