The smallest value of sum of reversed absolute values I want to find minimum of expression $$\frac{1}{|x_1 - x_2||x_1 - x_3|} + \frac{1}{|x_2 - x_1||x_2 - x_3|} + \frac{1}{|x_3 - x_1||x_3 - x_2|}$$
where $x_1, x_2, x_3 \in [-1, 1]$.
I'm struggeling with this problem a little bit. I tried some inequalities between means, but I'm not sure how I can exactly say what's the minimum value of it. Could you please give me a hint how it can be done?
 A: Since the expression $f(x_1,x_2,x_3)$ at hand is symmetric in the variables, we may assume without loss of generality that $x_1 < x_2 < x_3$. Note that in this case,
$$f(x_1,x_2,x_3) = \frac{2}{(x_3 - x_2)(x_2 - x_1)} \geqslant \frac{8}{(x_3 - x_1)^2} \geqslant 2,$$
where we have used the AM-GM inequality and the fact that $|x_3 - x_1| \leqslant 2$. Note that $2$ is indeed the minimum since $f(-1, 0, 1) = 2$.
A: This is a very interesting question. So, we are trying to minimize an expression:
$$
\min{\dfrac{1}{|x_1-x_2||x_1-x_3|}+\dfrac{1}{|x_2-x_1||x_2-x_3|}+\dfrac{1}{|x_3-x_1||x_3-x_1|}} 
$$
One way to minimize the expression is to minimize each of the terms individually, that is to say:
$$
\min (x+y+z) = \min(x)+\min(y)+\min(z)
$$
There are two differences in the denominator of $\dfrac{1}{|x_1-x_2||x_1-x_3|}$. Maximizing the denominator minimizies the fraction:
$$
min \dfrac{1}{a} = \dfrac{1}{\max a}
$$
Note: The maximum value of this denominator is 4 for the values $x_1 = 1, x_2 = -1, x_3 = -1$. However, we note that this leads to a cumbersome division by zero in the second and third terms. We note that $x_1\neq x_2 \neq x_3$ solves this problem.
If we pick the values of $x_1,x_2,\text{ and }x_3$ symmetrically, we can maximize each term:
$$
x_1 = -1,x_2 = 0,x_3 = 1
$$
Or any combination of these values.
This leads us to our answer:
$$
\begin{align}
& \dfrac{1}{|x_1-x_2||x_1-x_3|} & + & \dfrac{1}{|x_2-x_1||x_2-x_3|} & + & \dfrac{1}{|x_3-x_1||x_3-x_1|} \\
= & \dfrac{1}{|-1|\times|-2|} & + &\dfrac{1}{|1|\times|-1|} & + & \dfrac{1}{|2|\times|1|}\\
= &\dfrac{1}{2} & + & \dfrac{1}{1} & + & \dfrac{1}{2} 
\end{align}
= 2
$$
