jmc

We can group the factors in pairs: $$\begin{array}{rl}x(x+1)(x+2)(x+3)& =x(x+3)\cdot (x+1)(x+2)\\ & =(x^2+3x)(x^2+3x+2).\end{array}$$Let $y=x^2+3x+1.$ Then $$(x^2+3x)(x^2+3x+2)=(y-1)(y+1)=y^2-1\ge -1.$$Equality occurs when $y=x^2+3x+1=0,$ which has roots $x=\frac{-3\pm \sqrt{5}}{2}.$ Hence, the minimum value is $\boxed{-1}.$