Cho 2 số thực dương x, y thỏa mãn log3[(x + 1)(y + 1)]^y + 1 = 9 - (x - 1)(y + 1)

* Hướng dẫn giải

Với x, y > 0 ta có:

${\log }_3{\left[\left(x+1\right)\left(y+1\right)\right]}^{y+1}=9-\left(x-1\right)\left(y+1\right)\iff \left(y+1\right){\log }_3\left[\left(x+1\right)\left(y+1\right)\right]=9-\left(x-1\right)\left(y+1\right)$

$\iff {\log }_3\left(x+1\right)+{\log }_3\left(y+1\right)=\frac{9}{y+1}-x+1\iff {\log }_3\left(x+1\right)+\left(x+1\right)=2-{\log }_3\left(y+1\right)+\frac{9}{y+1}$

$\iff {\log }_3\left(x+1\right)+\left(x+1\right)={\log }_3\frac{9}{y+1}+\frac{9}{y+1}\left(1\right).$

Xét hàm số $f\left(t\right)={\log }_{3}t+t$ với t > 0

Ta có: $f\text{'}\left(t\right)=\frac{1}{t.\ln 3}+1>0,\forall t>0.$

$\Rightarrow$ Hàm số f(t) đồng biến trên khoảng $\left(0;+\infty \right).$

Khi đó: $\left(1\right)\iff f\left(x+1\right)=f\left(\frac{9}{y+1}\right)\iff x+1=\frac{9}{y+1}.$

Từ đó suy ra $P=x+2y=x+1+2y-1=\frac{9}{y+1}+2\left(y+1\right)-3\ge 2\sqrt{\frac{9}{y+1}.2\left(y+1\right)}-3=-3+6\sqrt{2}.$

Dấu “=” xảy ra $\iff \frac{9}{y+1}=2\left(y+1\right)\iff {\left(y+1\right)}^2=\frac{9}{2}\iff y=\frac{3\sqrt{2}}{2}-1\Rightarrow x=\frac{-25+27\sqrt{2}}{7}$ (thỏa mãn điều kiện x, y > 0).

Vậy $P_{\min }=-3+6\sqrt{2}$ khi $x=\frac{-25+27\sqrt{2}}{7};y=\frac{3\sqrt{2}}{2}-1.$

Chọn D.