Ta có:
$3a^2+b^2+3ab$
$=\dfrac{3}{4}a^2+\dfrac{3}{2}ab+\dfrac{3}{4}b^2+\dfrac{9}{4}a^2+\dfrac{3}{2}ab+\dfrac{1}{4}b^2$
$=\dfrac{3}{4}(a+b)^2+\dfrac{1}{4}(3a+b)^2\ge\dfrac{27}{4}$
Dấu bằng xảy ra khi và chỉ khi: $\left\{\begin{array}{l}a+b=3\\3a+b=0\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}a=-\dfrac{3}{2}\\b=\dfrac{9}{2}\end{array}\right.$