\(a;b>0\Rightarrow3a+2b+1>1\)
\(\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)\) đồng biến
Mà \(9a^2+b^2\ge2\sqrt{9a^2b^2}=6ab\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)\ge log_{3a+2b+1}\left(6ab+1\right)\)
\(\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)+log_{6ab+1}\left(3a+2b+1\right)\ge log_{3a+2b+1}\left(6ab+1\right)+log_{6ab+1}\left(3a+2b+1\right)\ge2\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}log_{6ab+1}\left(3a+2b+1\right)=1\\3a=b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6ab+1=3a+2b+1\\b=3a\end{matrix}\right.\)
\(\Rightarrow18a^2+1=3a+6a+1\)
\(\Leftrightarrow18a^2-9a=0\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{3}{2}\end{matrix}\right.\)
