Minimising expression with absolute values Let $x,y,z$ be distinct reals and consider the expression $$L=\frac{(|x|+|y|+|z|)^3}{|(x-y)(x-z)(y-z)|}$$ Find the minimum possible value of $L$ over all $(x,y,z)$. 
My work
Using a calculator, the answer seems to be $4$. One way of achieving this is when $x=1,y=-1,z=0$. I am not sure how to prove that this is the minimum, but the fact that $4$ is a square may lead to something, via the Cauchy-Schwarz inequality for example.
Also, I think we can assume $x\geq y\geq z$ because of something to do with symmetry? (Not sure on this). Then we can just consider lots of cases (E.g. $x\geq y\geq0\geq z$, etc.) and then simplify the absolute value expressions, making it easier to determine the minimum).  
My final thought is if there is a geometric solution. Since absolute values represent area, I'm wondering if we can consider certain points in the plane and use something like the triangle inequality to finish. This seems unlikely though, as $(x-y)(x-z)(y-z)$ doesn't seem to have a particularly nice geometric interpretation. 
EDIT: This question, coming from a contest, was implicitly designed not to have calculus-based solutions, so I'd appreciate if answers didn't use calculus.
 A: The expression for $L$ does not change if $x, y,z$ are permuted, so you may indeed assume that $x > y > z$.
With $a = x-y > 0$ and $b = y-z > 0$ we have
$$
    |x| + |y| + |z| \ge x-z = a+b 
$$
and
$$
  |(x-y)(x-z)(y-z)| = a(a+b)b
$$
and therefore
$$
L \ge \frac{(a+b)^3}{a(a+b)b} = \frac{(a+b)^2}{ab}  =  \frac{(a-b)^2}{ab} + 4 \ge 4\, .
$$
Equality holds exactly if $y=0$ and $a=b$, i.e. for $(x,y,z) = (a, 0, -a)$ with $a > 0$.
A: Assume that $x\gt y\gt z$ and also without loss of generality we can assume that $x\gt0$.
put: $x-a=y$  and  $x-a-b=z$ in which $a,b$ are positive numbers. The most important point is that such $a,b$ definitely exist. 
Now, in the first step assume that $y\geq0$ but $z\leq0$. So, we will have $a+b\geq x\geq a$ then we can rewrite the main inequality as below: $$\frac{((x)+(x-a)+(a+b-x))^3}{ab(a+b)}=\frac{(x+b)^3}{ab(a+b)}\geq\frac{(a+b)^3}{ab(a+b)}\geq\frac{(a+b)^2}{ab}\geq4$$
Notice $f(x)=\frac{(x+b)^3}{ab(a+b)}$ as a function of the variable $x$ is increasing in the $a+b\geq x\geq a$ range, while $a,b$ are considered to be constant.
Easily, you can check the other situations. For example, when $y\leq0$ and $z\leq0$.
4 is obtained when we put $x=a,y=0,z=-a$.
A: To find the minimum of a function, you have to find the derivative and then find the root of the derivative. For 3 variables it is essentially the same as for a single variable: Find the 3 partial derivatives 
$\frac{\partial L}{\partial x}$, $\frac{\partial L}{\partial y}$ , $\frac{\partial L}{\partial z}$ 
You will end up with 3 equations. Solve this system of equations for x,y,z. For this solution $L$ will be minimal.
To find the derivative of $|x|$ you can use $|x| = \sqrt{(x^2)}$. Therefore using the chain rule we have $(x)' = \frac{x}{|x|}$
