Cho các số thực $a,b\in \left( 1;3 \right]$ thỏa mãn $a<b$. Biết...

Ta có: $\forall b\in \left( 1;3 \right]$ : ${{b}^{2}}+9b-9\ge 3{{b}^{2}}$
Do đó: ${{\log }_{a}}\left( {{b}^{2}}+9b-9 \right)\ge {{\log }_{a}}\left( 3{{b}^{2}} \right)\ge {{\log }_{a}}\left( {{b}^{3}} \right)=3{{\log }_{a}}b$
Dấu "=" xảy ra $\Leftrightarrow b=3$
$\Rightarrow P\ge 3{{\log }_{a}}b+\dfrac{6}{{{\left( {{\log }_{a}}b-1 \right)}^{2}}}=3\left[ 1+\dfrac{{{\log }_{a}}b-1}{2}+\dfrac{{{\log }_{a}}b-1}{2}+\dfrac{2}{{{\left( {{\log }_{a}}b-1 \right)}^{2}}} \right]$
Theo BĐT Cô-si ta có:
$\dfrac{{{\log }_{a}}b-1}{2}+\dfrac{{{\log }_{a}}b-1}{2}+\dfrac{2}{{{\left( {{\log }_{a}}b-1 \right)}^{2}}}\ge 3\sqrt[3]{{{\left( \dfrac{{{\log }_{a}}b-1}{2} \right)}^{2}}.\dfrac{2}{{{\left( {{\log }_{a}}b-1 \right)}^{2}}}}\ge 3\sqrt[3]{\dfrac{1}{2}}$
$\Rightarrow P\ge 3\left[ 1+\dfrac{{{\log }_{a}}b-1}{2}+\dfrac{{{\log }_{a}}b-1}{2}+\dfrac{2}{{{\left( {{\log }_{a}}b-1 \right)}^{2}}} \right]\ge 3.\left[ 3\sqrt[3]{\dfrac{1}{2}}+1 \right]=9\sqrt[3]{\dfrac{1}{2}}+3$
Dấu "=" xảy ra $\Leftrightarrow \left\{ \begin{aligned}
& b=3 \\
& \dfrac{{{\log }_{a}}b-1}{2}=\dfrac{2}{{{\left( {{\log }_{a}}b-1 \right)}^{2}}} \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& b=3 \\
& {{\left( {{\log }_{a}}b-1 \right)}^{3}}=4 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& b=3 \\
& {{\log }_{a}}3-1=\sqrt[3]{4} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& b=3 \\
& 3={{a}^{1+\sqrt[3]{4}}} \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& b=3 \\
& {{3}^{\dfrac{1}{1+\sqrt[3]{4}}}}=a \\
\end{aligned} \right.$.
$\Rightarrow m=2;n=3\Rightarrow S=13$.