Cho hai số thực dương a, b thỏa mãn $\left( a+b \right)\left(...

Xét $\left( a+b \right)\left( 2a-ab+2b-2 \right)=3ab\Leftrightarrow 2{{a}^{2}}+2{{b}^{2}}-{{a}^{2}}b-a{{b}^{2}}+ab-2a-2b=0$
Vì $a,b$ dương, nên chia cho $ab$ ta được $\begin{aligned}
& 2\left( \dfrac{a}{b}+\dfrac{b}{a} \right)+1-\left( a+b \right)-2\dfrac{1}{b}-2\dfrac{1}{a}=0\Leftrightarrow 2\left( \dfrac{a}{b}+\dfrac{b}{a} \right)+1=\left( a+b \right)+2\left( \dfrac{1}{a}+\dfrac{1}{b} \right)\ge 2\sqrt{2.\left( a+b \right).\left( \dfrac{1}{a}+\dfrac{1}{b} \right)} \\
& \text{ }2\left( \dfrac{a}{b}+\dfrac{b}{a} \right)+1\ge 2\sqrt{2.\left( \dfrac{a}{b}+\dfrac{b}{a}+2 \right)} \\
\end{aligned}$
Suy ra $\dfrac{a}{b}+\dfrac{b}{a}\ge \dfrac{5}{2}$
Ta có $P=\dfrac{1}{{{a}^{3}}{{b}^{3}}}\left( {{a}^{6}}+{{b}^{6}} \right)-\dfrac{9}{4{{a}^{2}}{{b}^{2}}}\left( {{a}^{4}}+{{b}^{4}} \right)=\dfrac{{{a}^{3}}}{{{b}^{3}}}+\dfrac{{{b}^{3}}}{{{a}^{3}}}-\dfrac{9}{4}\left( \dfrac{{{a}^{2}}}{{{b}^{2}}}+\dfrac{{{b}^{2}}}{{{a}^{2}}} \right)\Leftrightarrow 4P=4\left( \dfrac{{{a}^{3}}}{{{b}^{3}}}+\dfrac{{{b}^{3}}}{{{a}^{3}}} \right)-9\left( \dfrac{{{a}^{2}}}{{{b}^{2}}}+\dfrac{{{b}^{2}}}{{{a}^{2}}} \right)$
Đặt $t=\dfrac{a}{b}+\dfrac{b}{a},t\ge \dfrac{5}{2}$ ta có được $4P=4({{t}^{3}}-3t)-9({{t}^{2}}-2)=4{{t}^{3}}-9{{t}^{2}}-12t+18$
Xét $f(t)=4{{t}^{3}}-9{{t}^{2}}-12t+18$ với $t\ge \dfrac{5}{2}$
$f'(t)=12{{t}^{2}}-18{{t}^{2}}-12\ge 0,\forall t\in \left[ \dfrac{5}{2}; \right.\left. +\infty \right)$ nên ${{f}_{\min }}(t)=f\left( \dfrac{5}{2} \right)=\dfrac{-23}{4}$
Do đó
$\begin{aligned}
& 4P=4({{t}^{3}}-3t)-9({{t}^{2}}-2)=4{{t}^{3}}-9{{t}^{2}}-12t+18\ge \dfrac{-23}{4} \\
& P\ge \dfrac{-23}{16} \\
\end{aligned}$
Dấu $''=''$ xảy ra khi $(a,b)=(2,1)\vee (1,2)$