\(P=\frac{9a^2+b^2+1}{4}+\frac{1}{\left(6ab+1\right)^2}\ge\frac{6ab+1}{4}+\frac{1}{\left(6ab+1\right)^2}\)
\(P\ge\frac{6ab+1}{8}+\frac{6ab+1}{8}+\frac{1}{\left(6ab+1\right)^2}\ge3\sqrt[3]{\frac{\left(6ab+1\right)^2}{64\left(6ab+1\right)^2}}=\frac{3}{4}\)
\(P_{min}=\frac{3}{4}\) khi \(\left\{{}\begin{matrix}9a^2=b^2\\\frac{6ab+1}{8}=\frac{1}{\left(6ab+1\right)^2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=3a\\ab=\frac{1}{6}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{\sqrt{18}}\\b=\frac{3}{\sqrt{18}}\end{matrix}\right.\)