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 $1\le a\le b\le c$.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $$(1+a+b)(1+a+c)(1+b+c)\geq9(ab+ac+bc)$$ or
$$1+2(a+b+c)+\sum_{cyc}(a^2+3ab)+\prod\limits_{cyc}(a+b)\geq9(ab+ac+bc)$$ or
$$1+6u+9u^2+3v^2+9uv^2-w^3\geq27v^2,$$ which is a linear inequality of $v^2$, which says that it remains to prove our inequality for the extremal value of $v^2$.
$a$, $b$ and $c$ are three non-negative roots of the equation 
$(x-a)(x-b)(x-c)=0$ or $3v^2x=-x^3+3ux^2+w^3$.
Thus, a line $y=3v^2x$ and a graph of $y=-x^3+3ux^2+w^3$ have three common points.
Hence, $v^2$ gets an extremal value, when a line $y=3v^2x$ is a tangent line to the graph of $y=-x^3+3ux^2+w^3$, which happens for equality case of two variables.
Let $b=a$.
Hence, we need to prove that
$$(2a+1)c^2+2(2a^2-6a+1)c+2a^3-4a^2+4a+1\geq0,$$ for which it's enough to probe that
$$(2a^2-6a+1)^2-(2a+1)(2a^3-4a^2+4a+1)\leq0,$$ which is $a(a-1)^2\geq0$.
Done!
