Cho $a,b>0$ thỏa mãn ${{\log }_{2a+3b+1}}\left(...

Với $a,b>0\Rightarrow \left\{ \begin{aligned}
& 25{{a}^{2}}+{{b}^{2}}+1>1 \\
& 2a+3b+1>1 \\
& 10ab+1>0 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{\log }_{2a+3b+1}}\left( 25{{a}^{2}}+{{b}^{2}}+1 \right)>0 \\
& {{\log }_{10ab+1}}\left( 2a+3b+1 \right)>0 \\
\end{aligned} \right.$
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 $\Leftrightarrow \left\{ \begin{aligned}
& 5a=b \\
& {{\log }_{2a+3b+1}}\left( 10ab+1 \right)=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 5a=b \\
& 10ab+1=2a+3b+1 \\
\end{aligned} \right.$
$\Rightarrow 50{{a}^{2}}=2a+15a\Rightarrow a=\dfrac{17}{50}\Rightarrow b=\dfrac{17}{10}$.