Cho a,b>0 thỏa mãn log(2a+3b+1)(25a^2+b^2+1)

Đáp án C

Với \(a,b > 0 \Rightarrow \left\{ \begin{array}{l}25{a^2} + {b^2} + 1 > 1\\2a + 3b + 1 > 1\\10ab + 1 > 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\log _{2a + 3b + 1}}\left( {25{a^2} + {b^2} + 1} \right) > 0\\{\log _{10ab + 1}}\left( {2a + 3b + 1} \right) > 0\end{array} \right.\)

Ta có \(P = {\log _{2a + 3b + 1}}\left( {25{a^2} + {b^2} + 1} \right) + {\log _{10ab + 1}}\left( {2a + 3b + 1} \right)\)

\( \ge {\log _{2a + 3b + 1}}\left( {10ab + 1} \right) + {\log _{10ab + 1}}\left( {2a + 3b + 1} \right)\)

\( \ge 2\sqrt {{{\log }_{2a + 3b + 1}}\left( {10ab + 1} \right).{{\log }_{10ab + 1}}\left( {2a + 3b + 1} \right)} = 2\).

Dấu “=” xảy ra \( \Leftrightarrow \left\{ \begin{array}{l}5a = b\\{\log _{2a + 3b + 1}}\left( {10ab + 1} \right) = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}5a = b\\10ab + 1 = 2a + 3b + 1\end{array} \right.\)

\( \Rightarrow 50{a^2} = 2a + 15a \Rightarrow a = \frac{{17}}{{50}} \Rightarrow b = \frac{{17}}{{10}}\).