Infimum of abc

Find the largest number $M$ such that for any given distinct number $a,b,c$ such that $0 < a,b,c < 1$ we always have: $$\frac{(a+b)(b+c)(c+a)}{|(a-b)(b-c)(c-a)|} > M$$

Solution

Clearly $M=1$ satisfies the condition. Now we prove that we can get the LHS arbitrarily close to 1.

Let $a$ be an arbitrary number < 1, $b = a/(1+x)$ and $c=a/(1+x+yx)$ for some really large numbers $x,y$. The LHS now becomes: $$\frac{(2+x)(2+(y+1)x)(2+(y+2)x)}{y(y+1)x^3}$$ $$=(\frac{2}{x}+1)(\frac{2}{x(y+1)} + 1)(\frac{2}{xy} + \frac{2}{y} + 1)$$ We can set $x,y$ large enough that the expression is as close as needed to 1.