57. Let $a=x^{2}, b=y^{2}, c=z^{2}$, then the problem reduces to proving under the condition $a, b, c \geqslant 0$
$$\begin{array}{c} \frac{a}{2 a+1}+\frac{b}{2 b+1}+\frac{c}{2 c+1} \leqslant \frac{b}{2 a+1}+\frac{c}{2 b+1}+\frac{a}{2 c+1} \\ (1) \Leftrightarrow 12 a b c+2(a b+b c+c a) \leqslant 4\left(a^{2} b+b^{2} c+c^{2} a\right)+2\left(a^{2}+b^{2}+c^{2}\right) \end{array}$$

By the AM-GM inequality, we have $a^{2} b+b^{2} c+c^{2} a \geqslant 3 \sqrt[3]{a^{2} b \cdot b^{2} c \cdot c^{2} a}=3 a b c, a^{2}+b^{2}+c^{2} \geqslant$ $a b+b c+c a$, so inequality (2) holds.