55. (2007.03.05) Simplify and prove: First prove: If $a^{2}+b^{2}+c^{2} \geqslant 3$, then

Because
$$\left(\sum a^{3}\right)^{2} \geqslant 3+2 \sum a^{4}$$
$$a^{2}+b^{2}+c^{2} \geqslant 3$$

So
$$3+2 \sum a^{4} \leqslant 3\left(\frac{\sum a^{2}}{3}\right)^{3}+2 \frac{\sum a^{2}}{3} \sum a^{4}$$

Therefore, it suffices to prove
$$\begin{array}{l} 9\left(\sum a^{3}\right)^{2} \geqslant\left(\sum a^{3}\right)^{3}+6 \sum a^{2} \cdot \sum a^{4} \Leftrightarrow \\ 2 \sum a^{6}+18 \sum b^{3} c^{3}-9 \sum b^{2} c^{2}\left(b^{2}+c^{2}\right)-6 a^{2} b^{2} c^{2} \geqslant 0 \Leftrightarrow \\ \sum\left(b^{4}-c^{4}\right)\left(b^{2}-c^{2}\right)-8 \sum b^{2} c^{2}(b-c)^{2}+\sum a^{2}(b c+c a+a b)(b-c)^{2} \geqslant 0 \Leftrightarrow \\ \sum\left[\left(b^{2}+c^{2}\right)(b+c)^{2}-8 b^{2} c^{2}+a^{2}(b c+c a+a b)\right](b-c)^{2} \geqslant 0 \end{array}$$

This inequality is clearly true, hence inequality (1) holds. Therefore,
$$\sum \frac{1}{x^{3}+y+z}=\sum \frac{1+y^{2}+z^{2}}{\left(x^{3}+y+z\right)\left(1+y^{2}+z^{2}\right)} \leqslant \frac{3+2 \sum x^{2}}{\left(\sum x^{\frac{3}{2}}\right)^{2}} \leqslant 1$$