129. Let $x=\frac{a}{b}, y=\frac{c}{a}, z=\frac{b}{c}$, then by the mean value inequality we get
$$\begin{aligned} (x+1)^{2}+2 y^{2}+1= & x^{2}+y^{2}+2 x+2 \geqslant 2 x y+2 x+2= \\ & \frac{2 c}{b}+\frac{2 a}{b}+2=\frac{2(a+b+c)}{b} \end{aligned}$$

So
$$\begin{array}{l} \frac{2}{(x+1)^{2}+2 y^{2}+1} \leqslant \frac{b}{a+b+c} \\ \frac{2}{(y+1)^{2}+2 z^{2}+1} \leqslant \frac{c}{a+b+c} \\ \frac{2}{(z+1)^{2}+2 x^{2}+1} \leqslant \frac{a}{a+b+c} \end{array}$$

Therefore
$$\frac{2}{(x+1)^{2}+2 y^{2}+1}+\frac{2}{(y+1)^{2}+2 z^{2}+1}+\frac{2}{(z+1)^{2}+2 x^{2}+1} \leqslant 1$$