Minimise it please

Since a, b and c are natural multiplication on the left is positive and is not zero. That means there are two negative multipliers, so let a < b < c.

Let b - a = x, c - b = y, where x and y are natural. Then c - a = x + y. Accordingly a = c - x - y, b = c - x:

x * y * (x + y) = (c - x - y) + (c - x) + c = a + b + c

x * y * (x + y) = 3c - 2x - y= a + b + c

Meaning : 3c = x * y * (x + y) + 2x + y. This expression must be divisible by 3 while it can only be when x and y are both divisible by 3.

Since we know that a + b + c = x * y * (x + y), that means the minimum is reached at minimal x and y values: x = 3, y = 3

then a = 15, b = 18, c = 21 and a + b + c = 54

Pigeonhole $\implies$ one of $(a-b), (b-c), (c-a)$ is even.

Let $x=a-b \quad y=b-c\,$

Then $\, c-a=-(x+y) \implies (a-b)(b-c)(c-a)=-x^2y-y^2x \qquad a+b+c=x+2y+3c$

$-x^2-y^2x \equiv x-y \pmod{3} \quad \text{either} \quad x^2\equiv 0 \pmod{3} \, \text{or} \, x^2\equiv 1 \pmod{3}$

If $x^2\equiv 1 \pmod{3} \, \text{ then} \, -y-y^2x\equiv x-y \implies y^2\equiv -1 \pmod{3} \text{\boxed{impossible}}$

So $3|x \implies 3|y \therefore 3|(x+y) \implies 3^3|xy(-x-y)$ which is even so the minimum is $2\times 27=\boxed{54}$