Cho $a>0$, $b>0$ thỏa mãn ${{\log }_{10a+3b+1}}\left(...

Từ giả thiết ta có $25{{a}^{2}}+{{b}^{2}}+1>0$, $10a+3b+1>0$, $10a+3b+1>1$, $10ab+1>1$. Áp dụng Cô-si, ta có $25{{a}^{2}}+{{b}^{2}}+1\ge 2\sqrt{25{{a}^{2}}{{b}^{2}}}+1=10ab+1$. Khi đó,
${{\log }_{10a+3b+1}}\left( 25{{a}^{2}}+{{b}^{2}}+1 \right)+{{\log }_{10ab+1}}\left( 10a+3b+1 \right)\ge {{\log }_{10a+3b+1}}\left( 10ab+1 \right)+{{\log }_{10ab+1}}\left( 10a+3b+1 \right)\ge 2$ l
(áp dụng Cô-si). Dấu $''=''$ xảy ra khi $\left\{ \begin{aligned}
& 5a=b \\
& {{\log }_{10a+3b+1}}\left( 10ab+1 \right)={{\log }_{10ab+1}}\left( 10a+3b+1 \right)=1 \\
\end{aligned} \right.$
Suy ra $\left\{ \begin{aligned}
& b=\dfrac{5}{2} \\
& a=\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow a+2b=\dfrac{11}{2}$