1. (1) If $x>0, y>0, z>0$, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $\frac{x y}{z}+\frac{y z}{x}+$ $\frac{z x}{y}$.
(2) Let $a, d \geqslant 0, b, c>0$, and $b+c \geqslant a+d$, find the minimum value of $\frac{b}{b+d}+\frac{c}{a+b}$.
1. (1) Let $x y=a z, y z=b x, \dot{x} z=c y$, then $a b=y^{2}, b c=z^{2}, c a=x^{2}, a b+b c+c a=1$. Therefore, $(a+b+c)^{2} \geqslant 3(a b+b c+c a)=3$, which means the minimum value of $a+b+c$ is $\sqrt{3}$, at this point $x=y=z=\frac{1}{\sqrt{3}}$.
(2) Since $b+c \geqslant a+d$, then $b+c \geqslant \frac{1}{2}(a+b+c+d)$. We can replace $a, b, c, d$ with $d$, $c, b, a$ respectively, and the expression $\frac{b}{c+d}+\frac{c}{a+b}$ remains unchanged. Without loss of generality, assume $a+b \geqslant c+d$, then $\frac{1}{c+d} \geqslant \frac{1}{a+b}$, thus
$$
\begin{aligned}
\frac{b}{c+d}+\frac{c}{a+b} & \geqslant \frac{1}{2} \frac{a+b+c+d}{c+d}-(c+d)\left(\frac{1}{c+d}-\frac{1}{a+b}\right) \\
& \geqslant 2 \sqrt{\frac{1}{2} \frac{a+b}{c+d} \cdot \frac{c+d}{a+b}}-\frac{1}{2}=\sqrt{2}-\frac{1}{2},
\end{aligned}
$$

When $a=\sqrt{2}+1, b=\sqrt{2}-1, c=2, d=0$, the minimum value $\sqrt{2}-\frac{1}{2}$ is achieved.