\(a^2+b^2+c^2-3\left(a+b-c\right)=a^2-4a+4+b^2-4b+4+c^2-c+\dfrac{1}{4}+a+b+4c-\left(4+4+\dfrac{1}{4}\right)=\left(a-2\right)^2+\left(b-2\right)^2+\left(c-\dfrac{1}{2}\right)^2+a+b+4c-\dfrac{33}{4}\ge3\sqrt[3]{4abc}-\dfrac{33}{4}=3\sqrt[3]{4.2}-\dfrac{33}{4}=-\dfrac{9}{4}\left(đpcm\right)\)
\(dấu"="\Leftrightarrow a=b=2;c=\dfrac{1}{2}\)