Cho a, b là hai số thực dương thoả mãn log 3a+2b+1 (9a^2 + b^2 + 1)

Ta có \({\left( {3a - b} \right)^2} \ge 0 \Leftrightarrow 9{a^2} - 6ab + {b^2} \ge 0 \Leftrightarrow 9{a^2} + {b^2} \ge 6ab\)

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

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

\( \ge 2\sqrt {{{\log }_{3a + 2b + 1}}\left( {6ab + 1} \right) \cdot {{\log }_{6ab + 1}}\left( {3a + 2b + 1} \right)}  = 2\).

Dấu bằng xảy ra khi và chỉ khi: \(\left\{ {\begin{array}{*{20}{l}}{3a - b = 0}\\{{{\log }_{3a + 2b + 1}}\left( {6ab + 1} \right) = 1}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{b = 3a}\\{6ab + 1 = 3a + 2b + 1}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{b = 3a}\\{6a \cdot 3a + 1 = 3a + 2.3a + 1}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{b = 3a}\\{18{a^2} - 9a = 0}\end{array}} \right.} \right.} \right.\)

\( \Leftrightarrow (a\,;\,\,b) = \left( {\frac{1}{2}\,;\,\,\frac{3}{2}} \right)\). Do đó \(a + 3b = \frac{1}{2} + 3 \cdot \frac{3}{2} = \frac{{10}}{2} = 5.\)

Đáp án: 5.