38. (1) By Cauchy-Schwarz inequality,
$$\begin{array}{l} \left(\frac{x}{a y+b z}+\frac{y}{a z+b x}+\frac{z}{a x+b y}\right)[x(a y+b z)+y(a z+b x)+z(a x+b y)] \geqslant \\ (x+y+z)^{2} \cdot(a+b)(x y+y z+z x)= \\ x(a y+b z)+y(a z+b x)+z(a x+b y) \\ \text { Also, }(x+y+z)^{2} \geqslant 3(x y+y z+z x), \text { so } \end{array}$$
$$\frac{x}{a y+b z}+\frac{y}{a z+b x}+\frac{z}{a x+b y} \geqslant \frac{3}{a+b}$$