Cho a > 0, b > 0 thỏa mãn ({log _{3a + 2b + 1}}left( {9{a^2} + {b^2} + 1} ight) + {log _{6ab + 1}}left( {3a + 2b + 1} ight) = 2

Ta có a > 0, b > 0 nên \(\left\{ \begin{array}{l}
3a + 2b + 1 > 1\\
9{a^2} + {b^2} + 1 > 1\\
6ab + 1 > 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\log _{3a + 2b + 1}}\left( {9{a^2} + {b^2} + 1} \right) > 0\\
{\log _{6ab + 1}}\left( {3a + 2b + 1} \right) > 0
\end{array} \right.\).

Áp dụng BĐT Cô-si cho hai số dương ta được

\({\log _{3a + 2b + 1}}\left( {9{a^2} + {b^2} + 1} \right) + {\log _{6ab + 1}}\left( {3a + 2b + 1} \right) \ge 2\sqrt {{{\log }_{3a + 2b + 1}}\left( {9{a^2} + {b^2} + 1} \right) + {{\log }_{6ab + 1}}\left( {3a + 2b + 1} \right)} \)

\(\begin{array}{l}
 \Leftrightarrow 2 \ge 2\sqrt {{{\log }_{6ab + 1}}\left( {9{a^2} + {b^2} + 1} \right)}  \Leftrightarrow {\log _{6ab + 1}}\left( {9{a^2} + {b^2} + 1} \right) \le 1 \Leftrightarrow 9{a^2} + {b^2} + 1 \le 6ab + 1\\
 \Leftrightarrow {\left( {3a - b} \right)^2} \le 0 \Leftrightarrow 3a = b
\end{array}\).

Vì dấu “=” đã xảy ra nên

\({\log _{3a + 2b + 1}}\left( {9{a^2} + {b^2} + 1} \right) = {\log _{6ab + 1}}\left( {3a + 2b + 1} \right) \Leftrightarrow {\log _{3b + 1}}\left( {2{b^2} + 1} \right) = {\log _{2{b^2} + 1}}\left( {3b + 1} \right)\)

\( \Leftrightarrow 2{b^2} + 1 = 3b + 1 \Leftrightarrow 2{b^2} - 3b = 0 \Leftrightarrow b = \frac{3}{2}\) (vì b > 0). Suy ra \(a = \frac{1}{2}\).

Vậy \(a + 2b = \frac{1}{2} + 3 = \frac{7}{2}\)