ABC Positive

Expanding and simplifying, we find that

$(1 - a)(1 - b)(1 - c) = (1 - a - b + ab)(1 - c) =$

$1 - a - b + ab - c + ac + bc + ab - abc = ab + ac + bc - abc$

since $a + b + c = 1$ . Now since $a,b,c$ are all positive, $ab,ac,bc$ will be as well. So by the AM-GM inequality we find that $a + b + c \ge 3(abc)^{\frac{1}{3}}$ and that $ab + ac + bc \ge 3(abc)^{\frac{2}{3}}$ .

Multiplying these two inequalities, we see that

$(a + b + c)(ab + ac + bc) \ge 9abc \Longrightarrow ab + ac + bc \ge 9abc$

since $a + b + c = 1$ . Thus

$(1 - a)(1 - b)(1 - c) = ab + ac + bc - abc \ge 9abc - abc = 8abc$ , and so $K = \boxed{8}$ .

(Note that equality holds when $a = b = c = \frac{1}{3}$ .)

this can be written as: $(a+b)(b+c)(c+a)$ WLOG, a≥b≥c. we have the sets (a,b,c) and (a,b,c). by reverse rearrangement inequality this is "random sum≥direct sum" $(a+b)(b+c)(c+a)≥(a+a)(b+b)(c+c)=\boxed{8}abc$ (equality achieved at $a=b=c=\dfrac{1}{3}$ .)

We can rewrite the given expression as $(a+b)(b+c)(c+a) \geq 2\sqrt{ab} \times 2\sqrt{bc} \times 2\sqrt{ca},\text{by A.M-G.M inequality}\\=8abc \Longrightarrow k=8.\\ \text{Equality occurs at}\ a=b=c=\dfrac{1}{3}$ .And done!