Áp dụng BĐT Cauchy-Schwarz:
\(\left(\frac{1}{2^2}+\frac{1}{\left(\sqrt{6}\right)^2}+\frac{1}{\left(\sqrt{3}\right)^2}\right)\left(\left(2x\right)^2+\left(y\sqrt{6}\right)^2+\left(z\sqrt{3}\right)^2\right)\ge\)
\(\left(\frac{1}{2}.2x+\frac{1}{\sqrt{6}}.y\sqrt{6}+\frac{1}{\sqrt{3}}.z\sqrt{3}\right)^2=\left(x+y+z\right)^2=3^2=9\)
\(\Rightarrow\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{3}\right)\left(4x^2+6y^2+3z^2\right)\ge9\)
\(\Leftrightarrow\frac{3}{4}A\ge9\Leftrightarrow A\ge12\)
Dấu = xảy ra \(\Leftrightarrow{\begin{cases}4x=6y=3z\\x+y+z=3\end{cases}\Leftrightarrow x=1,y=\frac{2}{3},z=\frac{4}{3}}\)
