Cho \(a > 0,\;b > 0\) thỏa mãn \({\log _{2a + 2b + 1}}\left( {4{a^2} + {b^2} + 1} \right) + {\log _{4ab + 1}}\left( {2a + 2b + 1} \right) = 2.\) Giá trị của \(a + 2b\) bằng:

Ta có: \({\log _{2a + 2b + 1}}\left( {4{a^2} + {b^2} + 1} \right) + {\log _{4ab + 1}}\left( {2a + 2b + 1} \right) = 2\)

\( \Leftrightarrow {\log _{2a + 2b + 1}}\left( {4{a^2} + {b^2} + 1} \right) + \dfrac{1}{{{{\log }_{2a + 2b + 1}}\left( {4ab + 1} \right)}} = 2.\)

Có: \({\left( {2a} \right)^2} + {b^2} \ge 2.2a.b \Leftrightarrow 4{a^2} + {b^2} \ge 4ab.\)

\( \Rightarrow 4{a^2} + {b^2} + 1 \ge 4ab + 1.\)

Dấu “=” xảy ra \( \Leftrightarrow 2a = b.\)

Theo giả thiết ta có: \(\left\{ \begin{array}{l}a > 0\\b > 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}2a + 2b + 1 > 1\\4ab + 1 > 1\end{array} \right. \Rightarrow \left\{ \begin{array}{ccccc}\log {_{2a + 2b + 1}}\left( {4{a^2} + {b^2} + 1} \right) \ge {\log _{2a + 2b + 1}}\left( {4ab + 1} \right)\\{\log _{4ab + 1}}\left( {2a + 2b + 1} \right) = \dfrac{1}{{{{\log }_{2a + 2b + 1}}\left( {4ab + 1} \right)}}\end{array} \right.\)

Áp dụng bất đẳng thức Cauchy ta có:

\({\log _{2a + 2b + 1}}\left( {4{a^2} + {b^2} + 1} \right) + \dfrac{1}{{{{\log }_{2a + 2b + 1}}\left( {4ab + 1} \right)}}\)\( \ge {\log _{2a + 2b + 1}}\left( {4ab + 1} \right) + \dfrac{1}{{{{\log }_{2a + 2b + 1}}\left( {4ab + 1} \right)}}\)  \( \ge 2.\sqrt {{{\log }_{2a + 2b + 1}}\left( {4ab + 1} \right).\dfrac{1}{{{{\log }_{2a + 2b + 1}}\left( {4ab + 1} \right)}}}  = 2.\)

Dấu “=” xảy ra \( \Leftrightarrow \left\{ \begin{array}{l}2a = b\\{\log _{2a + 2b + 1}}\left( {4ab + 1} \right) = \dfrac{1}{{{{\log }_{2a + 2b + 1}}\left( {4ab + 1} \right)}}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}2a = b\\\log _{3b + 1}^2\left( {2{b^2} + 1} \right) = 1\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}2a = b\\{\log _{3b + 1}}\left( {2{b^2} + 1} \right) = 1\end{array} \right.\)  (vì \({\log _{3b + 1}}\left( {2{b^2} + 1} \right) > 0\))

\( \Leftrightarrow \left\{ \begin{array}{l}b = 2a\\2{b^2} + 1 = 3b + 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 2a\\2{b^2} - 3b = 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}b = 2a\\\left[ \begin{array}{l}b = 0\;\;\left( {ktm} \right)\\b = \dfrac{3}{2}\;\;\left( {tm} \right)\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \dfrac{3}{4}\;\;\left( {tm} \right)\\b = \dfrac{3}{2}\end{array} \right.\)  

Vậy \(a + 2b = \dfrac{3}{4} + 3 = \dfrac{{15}}{4}.\)