Với lưu ý: \(log_36y=log_33.2y=1+log_32y\) và \(log_32xy=log_3x+log_32y\)
Đặt \(\left\{{}\begin{matrix}a=log_3x>0\\b=log_32y>log_32\end{matrix}\right.\)
\(\Rightarrow a\left(b+1\right)+2ab\left(3-a-b\right)=\dfrac{9}{2}\)
\(\Rightarrow7ab+a=\dfrac{9}{2}+2ab\left(a+b\right)\)
\(\Rightarrow7b+1=\dfrac{9}{2a}+2ab+2b^2\)
\(\Rightarrow-2b^2+7b+1=\dfrac{9}{2a}+2ab\ge2\sqrt{\dfrac{18ab}{2a}}=6\sqrt{b}\)
\(\Rightarrow-2b^2+7b-6\sqrt{b}+1\ge0\)
\(\Rightarrow\left(\sqrt{b}-1\right)^2\left(1-4\sqrt{b}-2b\right)\ge0\) (1)
Do \(b>log_32>log_3\sqrt{3}=\dfrac{1}{2}\Rightarrow1-4\sqrt{b}-2b< 0\)
Do đó (1) đúng khi và chỉ khi:
\(\sqrt{b}-1=0\Leftrightarrow b=1\) \(\Rightarrow a=\dfrac{3}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}log_3x=\dfrac{3}{2}\\log_32y=1\end{matrix}\right.\) \(\Rightarrow x+2y=\left(\dfrac{3}{2}\right)^3+3=\dfrac{51}{8}\) (chắc gần 7 nhất)
