$(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$ if $ab+bc+ca=3$

Question

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 ?

Answer 1

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.

Answer 2

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 .