$4 \cdot 27$ Set $M=\left\{(x, y) \mid x^{3}+8 y^{3}+6 x y \geqslant 1, \quad x, y \in R\right\}$, $D=\left\{(x, y) \mid x^{2}+y^{2} \leqslant t^{2}, \quad t \in R, \quad t \neq 0\right\}, D \cap M=\varnothing$, then
(A) $-1<t<1$.
(B) $-\frac{\sqrt{3}}{4}<t<\frac{\sqrt{3}}{4}$.
(C) $-\frac{\sqrt{5}}{5}<t<\frac{\sqrt{5}}{5}$.
(D) $-\frac{1}{2}<t<\frac{1}{2}$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] From $x^{3}+8 y^{3}+6 x y \geqslant 1$, we have
$$
x^{3}+(2 y)^{3}+(-1)^{3}-3 x(2 y)(-1) \geqslant 0 \text {. }
$$

which is $(x+2 y-1)\left[(x-2 y)^{2}+(2 y+1)^{2}+(x+1)^{2}\right] \geqslant 0$.
Since $\quad(x-2 y)^{2}+(2 y+1)^{2}+(x+1)^{2} \geqslant 0$,
$$
\therefore \quad x+2 y-1 \geqslant 0 \text {, }
$$

The distance from the origin $(0,0)$ to $x+2 y-1=0$ is $\frac{|0+2 \cdot 0-1|}{\sqrt{1^{2}+2^{2}}}=\frac{1}{\sqrt{5}}, \quad$ so $-\frac{1}{\sqrt{5}}<t<\frac{1}{\sqrt{5}}$. Therefore, the answer is $(C)$.