\(9=\left(\frac{1}{2}.2x+\frac{1}{\sqrt{6}}.\sqrt{6}y+\frac{1}{\sqrt{3}}\sqrt{3}z\right)^2\le\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{3}\right)\left(4x^2+6y^2+3z^2\right)\)
\(\Rightarrow A\ge\frac{9}{\frac{1}{4}+\frac{1}{6}+\frac{1}{3}}=12\)
\(A_{min}=12\) khi \(\left\{{}\begin{matrix}x+y+z=3\\4x=6y=3z\end{matrix}\right.\)