Từ: $3<ac,bc<6$ và $c\in[2;3]$ suy ra $1<a,b<3$.Lại có:
$2c(a^2+b^2)+b(ab+c)+c(ac+b)>b(b^2+c^2)+2ac(1+2b)$
$\Leftrightarrow (a-b)(2ac-c+b^2-2bc+c^2)>0$
$\Leftrightarrow a-b>0$ vì $2ac-c+b^2-2bc+c^2=2(a-1)c+(b-c)^2>0$
Đặt $x=a-b;y=b-1;z=3-a$ thì $x,y,z>0$ và $x+y+z=2$.
Khi đó: $P=2(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x})+9\sqrt[3]{xyz}$
Ta có:
$\dfrac{x}{y}+\dfrac{x}{y}+\dfrac{y}{z}\ge3\sqrt[3]{\dfrac{x^2}{yz}}=\dfrac{3x}{\sqrt[3]{xyz}}$
Tương tự: $\dfrac{y}{z}+\dfrac{y}{z}+\dfrac{z}{x}\ge\dfrac{3y}{\sqrt[3]{xyz}};\dfrac{z}{x}+\dfrac{z}{x}+\dfrac{x}{y}\ge\dfrac{3z}{\sqrt[3]{xyz}}$
Suy ra: $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\ge\dfrac{x+y+z}{\sqrt[3]{xyz}}=\dfrac{2}{\sqrt[3]{xyz}}$
$\Rightarrow P\ge\dfrac{4}{\sqrt[3]{xyz}}+9\sqrt[3]{xyz}\ge2\sqrt{\dfrac{4}{\sqrt[3]{xyz}}.9\sqrt[3]{xyz}}=12$
$\min P=12 \Leftrightarrow x=y=z=\dfrac{2}{3} \Leftrightarrow a=\dfrac{7}{3};b=\dfrac{5}{3}$