jmc

By AM-GM, $$x^2+1\ge 2x,$$so $$\frac{x^2+3x+1}{x}\ge \frac{5x}{x}=5.$$Likewise, $$\frac{y^2+3y+1}{y}\ge 5$$and $$\frac{z^2+3z+1}{z}\ge 5,$$so $$\frac{(x^2+3x+1)(y^2+3y+1)(z^2+3z+1)}{xyz}\ge 125.$$Equality occurs when $x=y=z=1,$ so the minimum value is $\boxed{125}.$