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

Ta có $a>0$, $b>0$ nên $\left\{ \begin{aligned}
& 3a+2b+1>1 \\
& 9{{a}^{2}}+{{b}^{2}}+1>1 \\
& 6ab+1>1 \\
\end{aligned} \right. $ $ \Rightarrow \left\{ \begin{aligned}
& {{\log }_{3a+2b+1}}\left( 9{{a}^{2}}+{{b}^{2}}+1 \right)>0 \\
& {{\log }_{6ab+1}}\left( 3a+2b+1 \right)>0 \\
\end{aligned} \right.$.
Áp dụng bất đẳng thức Cô-si cho hai số dương ta được:
${{\log }_{3a+2b+1}}\left( 9{{a}^{2}}+{{b}^{2}}+1 \right)+{{\log }_{6ab+1}}\left( 3a+2b+1 \right)\ge 2\sqrt{{{\log }_{3a+2b+1}}\left( 9{{a}^{2}}+{{b}^{2}}+1 \right).{{\log }_{6ab+1}}\left( 3a+2b+1 \right)}$
$\Leftrightarrow 2\ge 2\sqrt{{{\log }_{6ab+1}}\left( 9{{a}^{2}}+{{b}^{2}}+1 \right)}$ $\Leftrightarrow {{\log }_{6ab+1}}\left( 9{{a}^{2}}+{{b}^{2}}+1 \right)\le 1$
$\Leftrightarrow 9{{a}^{2}}+{{b}^{2}}+1\le 6ab+1$ $\Leftrightarrow {{\left( 3a-b \right)}^{2}}\le 0\Leftrightarrow 3a=b$.
Vì dấu “ $=$ ” đã xảy ra nên
${{\log }_{3a+2b+1}}\left( 9{{a}^{2}}+{{b}^{2}}+1 \right)={{\log }_{6ab+1}}\left( 3a+2b+1 \right)$ $\Leftrightarrow {{\log }_{3b+1}}\left( 2{{b}^{2}}+1 \right)={{\log }_{2{{b}^{2}}+1}}\left( 3b+1 \right)$
$\Leftrightarrow 2{{b}^{2}}+1=3b+1$ $\Leftrightarrow 2{{b}^{2}}-3b=0$ $\Leftrightarrow b=\dfrac{3}{2}$ (vì $b>0$ ) suy ra $a=\dfrac{1}{2}$.
Vậy $a+2b=\dfrac{1}{2}+3$ $=\dfrac{7}{2}$.