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 $a + 2b = \dfrac{3}{4} + 3 = \dfrac{{15}}{4}.$