jmc

By AM-HM, $$\frac{x_1+x_2+x_3+x_4}{4}\ge \frac{4}{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}},$$so $$\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\ge \frac{16}{x_1+x_2+x_3+x_4},$$for any positive real numbers $x_1,$ $x_2,$ $x_3,$ and $x_4.$

Taking $x_1=b+c+d,$ $x_2=a+c+d,$ $x_3=a+b+d,$ and $x_4=a+b+c,$ we find $$\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}\ge \frac{16}{3a+3b+3c+3d}=\frac{16}{3}.$$Since $a+b+c+d=1,$ we can write this as $$\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{a+c+d}+\frac{a+b+c+d}{a+b+d}+\frac{a+b+c+d}{a+b+c}\ge \frac{16}{3}.$$Then $$\frac{a}{b+c+d}+1+\frac{b}{a+c+d}+1+\frac{c}{a+b+d}+1+\frac{d}{a+b+c}+1\ge \frac{16}{3},$$so $$\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}\ge \frac{4}{3}.$$Equality occurs when $a=b=c=d=\frac{1}{4},$ so the minimum value is $\boxed{\frac{4}{3}}.$