Problem 1572

55. (1) When $k=1$, $(a+b+c)(a b+b c+c a)-\left(a^{3}+b^{3}+c^{3}\right)=a^{2}(b+$ $c-a)+b^{2}(c+a-b)+c^{2}(a+b-c)+3 a b c>0$, so $a^{3}+b^{3}+c^{3} & \frac{a^{3}+b^{3}+c^{3}}{(a+b+c)(a b+b c+c a)}= \\
& \frac{2\left(x^{3}+y^{3}+z^{3}\right)+3[x y(x+y)+y z(y+z)+z x(z+x)]}{2\left(x^{3}+y^{3}+z^{3}\right)+8[x y(x+y)+y z(y+z)+z x(z+x)]+18 x y z} \\
& k-1>-\frac{5[x y(x+y)+y z(y+z)+z x(z+x)]+18 x y z}{2\left(x^{3}+y^{3}+z^{3}\right)+8[x y(x+y)+y z(y+z)+z x(z+x)]+18 x y z}
\end{aligned} To ensure the above inequality always holds, $k \geqslant 1$. Another proof when $a=b=n, c=1$, \begin{aligned}
k> & \frac{2 n^{3}+1}{(2 n+1)\left(n^{2}+2 n\right)}=\frac{2 n^{3}+1}{2 n^{3}+5 n^{2}+2 n} \\
& \lim _{n \rightarrow \infty_{i}} \frac{2 n^{3}+1}{2 n^{3}+5 n^{2}+2 n}=1
\end{aligned}$$$$

Therefore, $k_{\text {min }}=1$.