Nice expression for minimum of three variables?

This probably isn't what you were thinking of, but if $a_1, ... a_n$ are non-negative, then

$$\min(a_1, ... a_n) = \lim_{k \to -\infty} \sqrt[k]{a_1^k + ... + a_n^k}$$

whereas

$$\max(a_1, ... a_n) = \lim_{k \to \infty} \sqrt[k]{a_1^k + ... + a_n^k}.$$

There are several applications of these identities (at least the second one), e.g. to functional analysis. They are also related to the way in which tropical arithmetic arises as a "limit" of ordinary arithmetic.

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 $$ \begin{align} \min(a,b,c) &=\frac{a}{2}\left(\frac{|c-a|+|a-b|}{|a-b|+|b-c|+|c-a|}\right)\\ &+\frac{b}{2}\left(\frac{|a-b|+|b-c|}{|a-b|+|b-c|+|c-a|}\right)\\ &+\frac{c}{2}\left(\frac{|b-c|+|c-a|}{|a-b|+|b-c|+|c-a|}\right)\\ &-\frac{|a-b|+|b-c|+|c-a|}{4}\tag{8} \end{align} $$

Here is a hint why there is no simple such formula:

In the case of two variables $a_i$ they are the zeros of the polynomial $$p(x):=(x-a_1)(x-a_2)=x^2 -(a_1+a_2)x + a_1 a_2\ .$$ Therefore by the formula for quadratic equations we have $$\min(a_1, a_2)\ =\ {a_1+a_2-\sqrt{(a_1+a_2)^2 -4a_1 a_2}\over 2} ={a_1+a_2\over2}-{|a_1-a_2|\over 2}\ .$$

Using the same idea with three variables $a_i$ we would have to look at the third degree polynomial $$q(x):=(x-a_1)(x-a_2)(x-a_3)$$ which has three real roots. Therefore we are in the "casus irreducibilis" of Cardano's formula, which can only be solved via complex numbers. Even if you would write everything out, you couldn't decide "by inspection" which root is the smallest.