57. If $x, y, z$ are all positive real numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $S=\frac{(z+1)^{2}}{2 x y z}$.
$\begin{array}{l}\text { 57. } S=\frac{(z+1)^{2}}{2 x y z} \geqslant \frac{(z+1)^{2}}{\left(x^{2}+y^{2}\right) z}=\frac{(z+1)^{2}}{\left(1-z^{2}\right) z}=\frac{z+1}{(1-z) z}=\frac{1-z+2 z}{(1-z) z}=\frac{1}{z}+\frac{2}{1-z}= \\ {[z+(1-z)] \cdot\left(\frac{1}{z}+\frac{2}{1-z}\right)=3+\frac{2 z}{1-z}+\frac{1-z}{z} \geqslant 3+2 \sqrt{\frac{2 z}{1-z} \cdot \frac{1-z}{z}}=3+2 \sqrt{2} .} \\ \text { Therefore, the minimum value of } S=\frac{(z+1)^{2}}{x y z} \text { is } 3+2 \sqrt{2} .\end{array}$