Cho a, b > 0 thỏa mãn \(...

A

- Hướng dẫn giải

Ta có \(16{{\rm{a}}^2} + {b^2} + 1 \ge 2\sqrt {16{{\rm{a}}^2}{b^2}} + 1 = 8{\rm{a}}b + 1\) 

Do đó:

\(lo{g_{4a + 5b + 1}}\left( {16{a^2} + {b^2} + 1} \right) + lo{g_{8ab + 1}}\left( {4a + 5b + 1} \right) \ge lo{g_{4a + 5b + 1}}\left( {8ab + 1} \right) + lo{g_{8ab + 1}}\left( {4a + 5b + 1} \right)\)

Mặt khác

\(\begin{array}{l}lo{g_{4a + 5b + 1}}\left( {8ab + 1} \right) + lo{g_{8ab + 1}}\left( {4a + 5b + 1} \right) = lo{g_{4a + 5b + 1}}\left( {8ab + 1} \right) + \frac{1}{{lo{g_{4a + 5b + 1}}\left( {8ab + 1} \right)}} \ge 2\\ \Rightarrow lo{g_{4a + 5b + 1}}\left( {16{a^2} + {b^2} + 1} \right) + lo{g_{8ab + 1}}\left( {4a + 5b + 1} \right) \ge 2\end{array}\)

Dấu “=” xảy ra \( \Leftrightarrow \left\{ \begin{array}{l}16{a^2} = {b^2}\\8ab + 1 = 4a + 5b + 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}4a = b\\2{b^2} + 1 = 6b + 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{3}{4}\\b = 3\end{array} \right.\).

Vậy \(a + 2b = \frac{{27}}{4}\).