Nice expression for minimum of three variables?

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

If you want a symmetric expression, you can take $\frac13 (\min(a,\min(b,c))+\min(b,\min(a,c))+\min(c,\min(a,b)))$. But if you rewrite it using that absolute-value trick, I still don't think it gives something that's "intuitively the minimum", I'm afraid...

Answer 4

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.