Cho $a>0,b>0$ thoả mãn ${{\log }_{4a+5b+1}}\left(...

Từ giả thiết suy ra ${{\log }_{4a+5b+1}}\left( 16{{a}^{2}}+{{b}^{2}}+1 \right)>0$ và ${{\log }_{8ab+1}}\left( 4a+5b+1 \right)>0$
Dễ dàng nhận thấy $16{{a}^{2}}+{{b}^{2}}+1\ge 8ab+1.$
Vì thế ${{\log }_{4a+5b+1}}\left( 16{{a}^{2}}+{{b}^{2}}+1 \right)+{{\log }_{8ab+1}}\left( 4a+5b+1 \right)$ $\ge {{\log }_{4a+5b+1}}\left( 8ab+1 \right)+{{\log }_{8ab+1}}\left( 4a+5b+1 \right)$
$\ge 2\sqrt{{{\log }_{4a+5b+1}}\left( 8ab+1 \right).{{\log }_{8ab+1}}\left( 4a+5b+1 \right)}=2$
Khi đó ${{\log }_{4a+5b+1}}\left( 16{{a}^{2}}+{{b}^{2}}+1 \right)+{{\log }_{8ab+1}}\left( 4a+5b+1 \right)=2$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{4a+5b+1}}\left( 8ab+1 \right)={{\log }_{8ab+1}}\left( 4a+5b+1 \right) \\
& b=4a \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{24a+1}}\left( 32{{a}^{2}}+1 \right)=1 \\
& b=4a \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 32{{a}^{2}}=24a \\
& b=4a \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{3}{4} \\
& b=3 \\
\end{aligned} \right.$
Vậy $a+2b=\dfrac{3}{4}+6=\dfrac{27}{4}.$
Chú ý: Với $a,b>1$ thì ${{\log }_{a}}b>0.$