Cho $\Large x,y>0$ thỏa mãn $\Large \log_{3}\left [ (x+1)(y+1) \right

Cho  $\Large x,y>0$ thỏa mãn $\Large \log_{3}\left [ (x+1)(y+1) \right ]^{y+1}=9-(x-1)(y+1)$. Tìm giá trị nhỏ nhất của biểu thức $\Large P=x+2y$

• A.

  A. $\Large P_{\min}=\dfrac{11}{2}$

• B.

  B. $\Large P_{\min}=\dfrac{27}{5}$

• C.

  C. $\Large P_{\min}=\dfrac{-1+6\sqrt{3}}{2}$

• D.

  D. $\Large P_{\min}=-3+6\sqrt{2}$

Chọn D

Ta có: 

$\Large \begin{align}&\log_{3}[(x+1)(y+1]^{y+1}=9-(x-1)(y+1)\\&\Leftrightarrow (y+1)\log_{3}[(x+1)(y+1)]=9-(x-1)(y+1)\\&\Leftrightarrow\log_{3}(x+1)+\log_{3}(y+1)=\dfrac{9}{y+1}-(x-1)\\&\Leftrightarrow\log_{3}(x+1)+x+1=2-\log_{3}(y+1)+\dfrac{9}{y+1}\\&\Leftrightarrow\log_{3}(x+1)+x+1=\log_{3}\left(\dfrac{9}{y+1}\right)+\dfrac{9}{y+1}\end{align}$

Xét hàm số $\Large f(t)=\log_{3}t+t, (t\in (0; +\infty))$ ta có $\Large f'(t)=\dfrac{1}{t.\ln3}+1>0, \forall t\in (0; +\infty) $

Do đó hàm số $\Large f(t)$ đồng biến trên khoảng $\Large (0; +\infty)$

Khi đó: 

$\Large f(x+1)=f\left(\dfrac{9}{y+1}\right)\Leftrightarrow x+1=\dfrac{9}{y+1}\Rightarrow P=\dfrac{9}{y+1}-1+2y=2(y+1)+\dfrac{9}{y+1}-3$

Mặt khác: $\Large 2(y+1)+\dfrac{9}{y+1}\geq 2\sqrt{2(y+1).\dfrac{9}{y+1}}=6\sqrt{2}\Rightarrow P_{\min}=6\sqrt{2}-3$