\(16a^2+b^2+1\ge8ab+1\)
\(\Rightarrow log_{4a+5b+1}\left(16a^2+b^2+1\right)+log_{8ab+1}\left(4a+5b+1\right)\ge2\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}b=4a\\4a+5b+1=8ab+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{4}\\b=3\end{matrix}\right.\)
Đáp án D