Cho hai số thực $a,b$ thỏa mãn $a>b>\dfrac{4}{3}$ và $16{{\log...

Ta có ${{b}^{3}}+16={{b}^{3}}+8+8\ge 3\sqrt[3]{64{{b}^{3}}}=12b\Rightarrow 12b-16\le {{b}^{3}}$
$\Rightarrow P=16.3-16{{\log }_{a}}\left( 12b-16 \right)+\dfrac{3}{{{\left( {{\log }_{a}}\dfrac{a}{b} \right)}^{2}}}\ge 48-16{{\log }_{a}}{{b}^{3}}+\dfrac{3}{{{\left( 1-{{\log }_{a}}b \right)}^{2}}}$
$\Rightarrow P\ge 48-48{{\log }_{a}}b+\dfrac{3}{{{\left( 1-{{\log }_{a}}b \right)}^{2}}}=48\left( 1-{{\log }_{a}}b \right)+\dfrac{3}{{{\left( 1-{{\log }_{a}}b \right)}^{2}}}$
Đặt $t=1-{{\log }_{a}}b>0\Rightarrow P\ge 48t+\dfrac{3}{{{t}^{2}}}=24t+24t+\dfrac{3}{{{t}^{2}}}\ge 3\sqrt[3]{24t.24t.\dfrac{3}{{{t}^{2}}}}=36$.
Dấu xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& b=2 \\
& 24t=\dfrac{3}{{{t}^{2}}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=2 \\
& t=\dfrac{1}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=2 \\
& b=\sqrt{a} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=2 \\
& a=4 \\
\end{aligned} \right.\Rightarrow a+b=6$.