57. Since $x y z=1, \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geqslant x+y+z \Leftrightarrow(x-1)(y-1)(z-1) \leqslant 0$. For any positive integer $k, t-1$ has the same sign as $t^{k}-1$, so
$$\begin{array}{l} \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geqslant x+y+z \Leftrightarrow(x-1)(y-1)(z-1) \leqslant 0 \Leftrightarrow \\ \left(x^{k}-1\right)\left(y^{k}-1\right)\left(z^{k}-1\right) \leqslant 0 \Leftrightarrow \\ \frac{1}{x^{k}}+\frac{1}{y^{k}}+\frac{1}{z^{k}} \geqslant x^{k}+y^{k}+z^{k} \end{array}$$