# If $(a+b)(a+c)(b+c)=8$ then $\prod\limits_{cyc}(2a+bc)\leq27$

Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that: $$(2a+bc)(2b+ac)(2c+ab)\leq27$$

My trying:

We need to prove that $$8abc+a^2b^2c^2+\sum_{cyc}(4a^2b^2+2a^3bc)\leq27$$ or $$abc\prod_{cyc}(a+b)+a^2b^2c^2+\sum_{cyc}a^2b^2\sqrt[3]{\prod_{cyc}(a+b)^2}+abc\sum_{cyc}a^2\sqrt[3]{\prod_{cyc}(a+b)}\leq\frac{27}{64}\prod_{cyc}(a+b)^2.$$ We made a homogenization! But what is the rest? Thank you!

let : $p= a+b+c, \ q= ab+bc+ca, \ r=abc$

then : $pq=8+r \ (\ p\ge 3, \ 0\le q\le 3)$

$$q^2\ge 3pr \Rightarrow q \le \dfrac{3p^2-\sqrt{9p^4-96p}}{2}$$

$$by \ Shur : q^3+9r \ge 4pq \Rightarrow q \ge \dfrac{72-p^3}{5p}$$

$$f(p,q) = (p-2)^2q^2+2(p^2-4)(p-4)q+16p(4-p) \le 27$$

So we only need to consider the inequality in 2 cases :

1. $f\left(p, \dfrac{3p^2-\sqrt{9p^4-96p}}{2}\right) \le 27, \ p\ge 3 \Leftrightarrow$

$$(p-3)(243p^5-465p^4-559p^3-3109p^2+8849p+243) \ge 0$$

1. $f\left(p, \dfrac{72-p^3}{5p}\right)\le 27, \ 3\le p \le 4 \Leftrightarrow$

$$(p-3)(p^7-11p^6+11p^5-71p^4+523p^3-287p^2+768p-6912) \le 0 , \ 3\le p \le 4$$