\[(a-1)(b-1)(c-1)\geq 0\leftrightarrow (ab-a-b+1)(c-1)\geq 0\leftrightarrow abc-ab-ac+a-bc+b+c-1\geq 0\leftrightarrow abc\geq ab+bc+ac-5\]

\[(3-a)(3-b)(3-c)\geq 0\leftrightarrow (9-3b-3a+ab)(3-c)\geq 0\Leftrightarrow 27-9c-9b-9a+3ac+3ab+3bc-abc\geq 0\leftrightarrow 3ab+3bc+3ac\geq 27+abc\geq ab+bc+ac-5+27\leftrightarrow ab+bc+ac\geq 11\]

\[\to P=\frac{ a^2b^2+12abc+72}{ab+bc+ac}-\frac{1}{2}abc\leq \frac{(ab+bc+ac)^2-12abc+12abc+72}{ab+bc+ac}-\frac{1}{2}(ab+bc+ac-5)=\frac{ab+bc+ac}{2}+\frac{72}{ab+bc+ac}+\frac{5}{2}=\frac{z}{2}+\frac{72}{z}+\frac{5}{2}(z=ab+bc+ac;z\geq 11)\]

Có: $z=ab+bc+ca\leq \dfrac{(a+b+c)^{2}}{3}=12\rightarrow 11 \leq z\leq 12$

Xét hiệu: $\dfrac{z}{2}+\frac{72}{z}+\frac{5}{2}-\frac{160}{11}=\frac{(11z-144)(z-11)}{22z}\leq 0$

$\rightarrow Max=\dfrac{160}{11}$

Đẳng thức xảy ra $↔(a;b;c)=(1;2;3)$ và các hoán vị

Hoặc bạn khảo cách giải sau:

\[(a-1)(b-1)(c-1)\geq 0\leftrightarrow (ab-a-b+1)(c-1)\geq 0\leftrightarrow abc-ab-ac+a-bc+b+c-1\geq 0\leftrightarrow abc\geq ab+bc+ac-5\]

\[(3-a)(3-b)(3-c)\geq 0\Leftrightarrow (9-3b-3a+ab)(3-c)\geq 0\Leftrightarrow 27-9c-9b-9a+3ac+3ab+3bc-abc\geq 0\Leftrightarrow 3ab+3bc+3ac\geq 27+abc\geq ab+bc+ac-5+27\Leftrightarrow ab+bc+ac\geq 11\]

\[\to P=\frac{ a^2b^2+12abc+72}{ab+bc+ac}-\frac{1}{2}abc\leq \frac{(ab+bc+ac)^2-12abc+12abc+72}{ab+bc+ac}-\frac{1}{2}(ab+bc+ac-5)=\frac{ab+bc+ac}{2}+\frac{72}{ab+bc+ac}+\frac{5}{2}=\frac{z}{2}+\frac{72}{z}+\frac{5}{2}(z=ab+bc+ac;z\geq 11)\]

Có: $z=ab+bc+ca\leq \dfrac{(a+b+c)^{2}}{3}=12\rightarrow 11 \leq z\leq 12$

Xét $\dfrac{z}{2}+\frac{72}{z}+\frac{5}{2}-\frac{160}{11}=\frac{(11z-144)(z-11)}{22z}\leq 0$

$\rightarrow Max P =\dfrac{160}{11}$

Vậy $ Max P =\dfrac{160}{11}$

Chúc bn hok tốt !!!!