Cho $a>0, b>0$ thỏa mãn ${{\log }_{4a+5b+1}}\left(16{{a}^{2}}+{{b}^{2}}+1 \right)+{{\log }_{8ab+1}}\left(4a+5b+1 \right)=2$. Giá trị của $a+2b$ bằng:

Ta có: $16{{a}^{2}}+{{b}^{2}}+1\ge 2\sqrt{16{{a}^{2}}{{b}^{2}}}+1=8ab+1$ do đó:
${{\log }_{4a+5b+1}}\left(16{{a}^{2}}+{{b}^{2}}+1 \right)+{{\log }_{8ab+1}}\left(4a+5b+1 \right)$
$\ge {{\log }_{4a+5b+1}}\left(8ab+1 \right)+{{\log }_{8ab+1}}\left(4a+5b+1 \right)$
$={{\log }_{4a+5b+1}}\left(8ab+1 \right)+\frac{1}{{{\log }_{4a+5b+1}}\left(8ab+1 \right)}$
$\ge 2\sqrt{{{\log }_{4a+5b+1}}\left(8ab+1 \right).\frac{1}{{{\log }_{4a+5b+1}}\left(8ab+1 \right)}}=2$
Dấu $''=''$ xảy ra khi và chỉ khi:
$\left\{ \begin{aligned}
& 16{{a}^{2}}={{b}^{2}} \\
& 8ab+1=4a+5b+1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 4a=b \\
& 2{{b}^{2}}+1=6b+1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\frac{3}{4} \\
& b=3 \\
\end{aligned} \right.$
Vậy $a+2b=\frac{27}{4}$.