jmc

By AM-GM, $$1+a=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+a\ge 4\sqrt[4]{\frac{1}{3^3}\cdot a}=4\sqrt[4]{\frac{a}{27}}.$$Similarly, $$\begin{array}{rl}1+b& \ge 4\sqrt[4]{\frac{b}{27}},\\ 1+c& \ge 4\sqrt[4]{\frac{c}{27}},\\ 1+d& \ge 4\sqrt[4]{\frac{d}{27}}.\end{array}$$Also by AM-GM, $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge 4\sqrt[4]{\frac{1}{abcd}}.$$Multiplying all these inequalities, we get $$\begin{array}{rl}(1+a)(1+b)(1+c)(1+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)& \ge 4\sqrt[4]{\frac{a}{27}}\cdot 4\sqrt[4]{\frac{b}{27}}\cdot 4\sqrt[4]{\frac{c}{27}}\cdot 4\sqrt[4]{\frac{d}{27}}\cdot 4\sqrt[4]{\frac{1}{abcd}}\\ & =\frac{1024}{27}.\end{array}$$Equality occurs when $a=b=c=d=\frac{1}{3},$ so the minimum value is $\boxed{\frac{1024}{27}}.$