<s>[INST] $(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$ if $ab+bc+ca=3$ Let $a,b,c$ be real positive number, $ab+bc+ca=3$.
Prove that $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$$
My attempt: $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 3\sqrt[3]{(a+b)(a+bc)(b+c)(b+ca)(c+a)(c+ab)}$$
$$P \ge 3\sqrt[3]{\frac{8}{9}\cdot(a+b+c)(ab+bc+ca)(a+bc)(b+ca)(c+ab)}$$
$$P \ge3\sqrt[3]{\frac{8}{3}\cdot(a+b+c)(a+bc)(b+ca)(c+ab)} $$
I can't prove $(a+bc)(b+ca)(c+ab)\ge 8$
Could you help me ?
 [/INST] Your last inequlaity is equivalent to
$$
a^2b^2c^2+abc(a^2+b^2+c^2)+a^2b^2+b^2c^2+c^2a^2+abc\ge 8.
$$
Since
$$
a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=(a+b+c)^2-6~\text{and}~
\\
a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=9-2abc(a+b+c),
$$
the previous inequlaity is equivalent to
$$
a^2b^2c^2+abc(a+b+c)^2-6abc+9-2abc(a+b+c)+abc\ge 8,
$$
or
$$
a^2b^2c^2+abc((a+b+c)^2-2(a+b+c))+1\ge 5abc.
$$
Now note that $(a+b+c)^2\ge 3(ab+bc+ca)=9$, so $a+b+c\ge 3$ and hence,
$$
(a+b+c)^2-2(a+b+c)=(a+b+c-1)^2-1\ge 3.
$$
Thus, it's sufficient to prove that
$$
a^2b^2c^2+3abc+1\ge 5abc\Leftrightarrow (abc-1)^2\ge 0,
$$
which is true.
</s>