Take $x=y=1$ to get $6 \geqslant 2 k \Rightarrow k \leqslant 3$. Now we prove: $4 x^{2}+y^{2}+1 \geqslant 3 x(y+1)$ holds for all integers. Since $4 x^{2}-3 x+1+y^{2}-3 x y=\frac{7}{4} x^{2}-3 x+1+\left(y-\frac{3}{2} x\right)^{2}$, note that the quadratic function $f(x)=\frac{7}{4} x^{2}-3 x+1$ has its axis of symmetry at $x=\frac{6}{7}$, and $f(0)=1>0$, $f(2)=2>0$, which means that except for $x=1$, $f(x)>0$ holds for all other integer values of $x$. When $x=1$, note that $y$ is an integer, then $4 x^{2}-3 x+1+y^{2}-3 x y=-\frac{1}{4}+\left(y-\frac{3}{2}\right)^{2}$ $\geqslant-\frac{1}{4}+\frac{1}{4}=0$, thus proving that $4 x^{2}-3 x+1+y^{2}-3 x y \geqslant 0$ holds for any integer $x$. In summary, the maximum constant $k=3$.
11. If for any integers $x, y$, the inequality $4 x^{2}+y^{2}+1 \geqslant k x(y+1)$ always holds, find the maximum constant $k$.