Ta có: \(a^2+b^2+c^2+4\le ab+3b+2c\)
Hay: \(\left(a^2-ab+\frac{b^2}{4}\right)+\left(\frac{3b^2}{4}-3b+3\right)+\left(c^2-2c+1\right)\le0\)
\(\Leftrightarrow\left(a-\frac{b}{2}\right)^2+3\left(\frac{b}{2}-1\right)^2+\left(c-1\right)^2\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}c-1=0\\\frac{b}{2}-1=0\\a-\frac{b}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=1\end{matrix}\right.\)
Vậy .....................