\(\dfrac{1}{3x^2+y^2}+\dfrac{2}{y^2+3xy}=\dfrac{1}{3x^2+y^2}+\dfrac{4}{2y^2+6xy}\ge\dfrac{\left(1+2\right)^2}{3x^2+y^2+2y^2+6xy}=\dfrac{9}{3\left(x^2+2xy+y^2\right)}=\dfrac{3}{\left(x+y\right)^2}\ge\dfrac{3}{1}=3\)-Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

