Cho các số thực $a,b$ thỏa mãn điều kiện $0<b<a<1$. Tìm giá trị...

Ta có $9{{b}^{2}}-12b+4={{\left( 3b-2 \right)}^{2}}\ge 0\Rightarrow \dfrac{4\left( 3b-1 \right)}{9}\le {{b}^{2}}$
$\Rightarrow P\ge {{\log }_{a}}{{b}^{2}}+8\log _{\dfrac{b}{a}}^{2}a=2{{\log }_{a}}b+8{{\left( \dfrac{1}{{{\log }_{a}}\dfrac{b}{a}} \right)}^{2}}=2{{\log }_{a}}b+\dfrac{8}{{{\left( {{\log }_{a}}b-1 \right)}^{2}}}.$
Đặt $t={{\log }_{a}}b-1>0\Rightarrow P\ge 2\left( t+1 \right)+\dfrac{8}{{{t}^{2}}}=t+t+\dfrac{8}{{{t}^{2}}}\ge 3\sqrt[3]{t.t.\dfrac{8}{{{t}^{2}}}}+2=8.$
Dấu "=" xảy ra $\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
b=\dfrac{2}{3} \\
t=\dfrac{8}{{{t}^{2}}} \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
b=\dfrac{2}{3} \\
t=2 \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
b=\dfrac{2}{3} \\
b={{a}^{3}} \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
b=\dfrac{2}{3} \\
a=\sqrt[3]{\dfrac{2}{3}} \\
\end{array} \right.$