Cho 2 số thực dương $x,y$ thỏa mãn ${{\log }_{3}}{{\left[ \left(x+1 \right)\left(y+1 \right) \right]}^{y+1}}=9-\left(x-1 \right)\left( y+1...

Với $x,y>0$ ta có:
${{\log }_{3}}{{\left[ \left( x+1 \right)\left( y+1 \right) \right]}^{y+1}}=9-\left( x-1 \right)\left( y+1 \right)\Leftrightarrow \left( y+1 \right){{\log }_{3}}\left[ \left( x+1 \right)\left( y+1 \right) \right]=9-\left( x-1 \right)\left( y+1 \right)$
$\Leftrightarrow {{\log }_{3}}\left( x+1 \right)+{{\log }_{3}}\left( y+1 \right)=\dfrac{9}{y+1}-x+1\Leftrightarrow {{\log }_{3}}\left( x+1 \right)+\left( x+1 \right)=2-{{\log }_{3}}\left( y+1 \right)+\dfrac{9}{y+1}$
$\Leftrightarrow {{\log }_{3}}\left( x+1 \right)+\left( x+1 \right)={{\log }_{3}}\dfrac{9}{y+1}+\dfrac{9}{y+1}\left( 1 \right).$
Xét hàm số $f\left( t \right)={{\log }_{3}}t+t$ với $t>0.$
Ta có: $f'\left( t \right)=\dfrac{1}{t.\ln 3}+1>0,\forall t>0.$
$\Rightarrow $ Hàm số $f\left( t \right)$ đồng biến trên khoảng $\left( 0;+\infty \right).$
Khi đó: $\left( 1 \right)\Leftrightarrow f\left( x+1 \right)=f\left( \dfrac{9}{y+1} \right)\Leftrightarrow x+1=\dfrac{9}{y+1}.$
Từ đó suy ra $P=x+2y=x+1+2y-1=\dfrac{9}{y+1}+2\left( y+1 \right)-3\ge 2\sqrt{\dfrac{9}{y+1}.2\left( y+1 \right)}-3=-3+6\sqrt{2}.$
Dấu "=" xảy ra $\Leftrightarrow \dfrac{9}{y+1}=2\left( y+1 \right)\Leftrightarrow {{\left( y+1 \right)}^{2}}=\dfrac{9}{2}\Leftrightarrow y=\dfrac{3\sqrt{2}}{2}-1\Rightarrow x=\dfrac{-25+27\sqrt{2}}{7}$ (thỏa mãn điều kiện $x,y>0$ ).
Vậy ${{P}_{\min }}=-3+6\sqrt{2}$ khi $x=\dfrac{-25+27\sqrt{2}}{7};y=\dfrac{3\sqrt{2}}{2}-1.$