Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$ How one can prove the following.
Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:
$(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$
WLOG one can assume that $0\le a\le b\le c$.
It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le a\le b\le c$.

$a=0, b=0, c=2 \ \implies \ 1 \cdot 3 \cdot 3 \ge 3 \cdot 2^2$, not true.
