Giải:
Vì \(x,y,z\) là các số nguyên nên:
\(x^2+y^2+z^2\le xy+3y+2z-3\)
\(\Leftrightarrow x^2+y^2+z^2-xy-3y-2z+3\le0\)
\(\Leftrightarrow\left(x^2-xy+\dfrac{y^2}{4}\right)+\left(\dfrac{3y^2}{4}-3y+3\right)+\left(z^2-2z+1\right)\le0\)
\(\Leftrightarrow\left(x-\dfrac{y}{2}\right)^2+3\left(\dfrac{y}{2}-1\right)^2+\left(z-1\right)^2\le0\)
Mà \(\left(x-\dfrac{y}{2}\right)^2+3\left(\dfrac{y}{2}-1\right)^2+\left(z-1\right)^2\le0\forall x,y\in R\)
\(\Leftrightarrow\left(x-\dfrac{y}{2}\right)^2+3\left(\dfrac{y}{2}-1\right)^2+\left(z-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{y}{2}=0\\\dfrac{y}{2}-1=0\\z-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=1\end{matrix}\right.\)
Vậy các số nguyên \(x,y,z\) phải tìm là: \(\left\{{}\begin{matrix}x=1\\y=2\\z=1\end{matrix}\right.\)