4. (2007.02.26) Prove briefly: If $x, y, z \in \mathbf{R}^{-}$, then $x y z \geqslant \prod(-x+y+z)$, i.e., $\sum x^{3}+3 x y z \geqslant \sum y z(y+z)$. The following will use this inequality. Since
$$\begin{aligned} 9 \sum b^{4} c^{4}= & 9 \sum\left(b^{3} c^{3} \cdot b c\right) \leqslant 3 \sum b^{3} c^{3}\left(b^{3}+c^{3}+1\right)= \\ & 3 \sum b^{3} c^{3}\left(b^{3}+c^{3}\right)+3 \sum b^{3} c^{3}= \\ & \sum b^{3} c^{3}\left(b^{3}+c^{3}\right)+2 \sum b^{3} c^{3}\left(b^{3}+c^{3}\right)+3 \sum b^{3} c^{3} \leqslant \\ & \left(\sum a^{9}+3 a^{3} b^{3}+2 \sum b^{3} c^{3} c b^{3}+c^{3}\right)+3 \sum b^{3} c^{3}= \\ & \sum a^{3}\left(\sum a^{6}+\sum b^{3} c^{3}\right)+3 \sum b^{3} c^{3}= \\ & \sum a^{3} \cdot\left(\sum a^{6}+2 \sum b^{3} c^{3}\right)=\left(\sum a^{3}\right)^{3}=27 \end{aligned}$$
(Note that $\sum a^{2}=3$).
Note: If $a^{3}+b^{3}+c^{3} \leqslant 3 \lambda^{3}$, then $3 \sum b^{4} c^{4} \leqslant \lambda^{2} \cdot\left(\sum a^{3}\right)^{2}$. The proof is similar to the above.