<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] Hints :
We have :
$$P=(a+b)(b+bc) + (b+c)(c+ca) + (c+a)(a+ab) \ge 12$$
Using :$$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0$$
Now we use Cauchy-Schwarz inequality to get :
$$P\geq \sum_{cyc}^{}(a+\sqrt{abc})^{2}\geq 12$$
Now expand and use the substitution $3u=a+b+c$,$3v^2=ab+bc+ca$ and $w^3=abc$
Conclude using uvw's method .
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