Xét các số thực $a,b$ thỏa mãn điều kiện $\dfrac{1}{3}<b<a<1$. Tìm...

Ta có $\dfrac{3b-1}{4}\le {{b}^{3}}\Leftrightarrow 4{{b}^{3}}-3b+1\ge 0\Leftrightarrow \left( b+1 \right)\left( 4{{b}^{2}}-4b+1 \right)\ge 0$
$\Leftrightarrow \left( b+1 \right){{\left( 2b-1 \right)}^{2}}\ge 0$ luôn đúng với $\dfrac{1}{3}<b<1.$
$\Rightarrow {{\log }_{a}}\left( \dfrac{3b-1}{4} \right)\ge {{\log }_{a}}{{b}^{3}}$ (vì $a<1$ ) $\Rightarrow {{\log }_{a}}\left( \dfrac{3b-1}{4} \right)\ge 3{{\log }_{a}}b$.
Biến đổi ${{\log }_{\dfrac{b}{a}}}a=\dfrac{1}{{{\log }_{a}}\dfrac{b}{a}}=\dfrac{1}{{{\log }_{a}}b-1}$
$\Rightarrow P\ge 3{{\log }_{a}}b+\dfrac{12}{{{\left( {{\log }_{a}}b-1 \right)}^{2}}}-3=3\left( {{\log }_{a}}b-1 \right)+\dfrac{12}{{{\left( {{\log }_{a}}b-1 \right)}^{2}}}$.
Bài ra $\dfrac{1}{3}<b<a<1\Rightarrow {{\log }_{a}}b>1$.
Đặt $t={{\log }_{a}}b-1>0\Rightarrow P\ge 3t+\dfrac{12}{{{t}^{2}}}=\dfrac{3t}{2}+\dfrac{3t}{2}+\dfrac{12}{{{t}^{2}}}\ge 3.\sqrt{\dfrac{3t}{2}.\dfrac{3t}{2}.\dfrac{12}{{{t}^{2}}}}=9$.
Dấu "=" xảy ra $\left\{ \begin{aligned}
& b=\dfrac{1}{2} \\
& \dfrac{3t}{2}=\dfrac{12}{{{t}^{2}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{1}{2} \\
& t=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{1}{2} \\
& b={{a}^{3}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{1}{2} \\
& a=\dfrac{1}{\sqrt[3]{2}} \\
\end{aligned} \right.$.