Who's up to the challenge? 76

Once you know the answer, the proof is pretty simple using the AM-GM inequality:

$(1+a)(1+b)(1+c)(1+d)\left(\dfrac{1}{a}+ \dfrac{1}{b} + \dfrac{1}{c} +\dfrac{1}{d}\right)$ $= \dfrac{1}{4}\left(\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{4}{3} + 4a\right)\dfrac{1}{4}\left(\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{4}{3} + 4b\right)\dfrac{1}{4}\left(\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{4}{3} + 4c\right)\dfrac{1}{4}\left(\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{4}{3} + 4d\right)\dfrac{1}{4}\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c} + \dfrac{4}{d}\right)$ $\ge \sqrt[4]{ \left(\dfrac{4}{3}\right)^3 4a \left(\dfrac{4}{3}\right)^3 4b \left(\dfrac{4}{3}\right)^3 4c \left(\dfrac{4}{3}\right)^3 4d \dfrac{4}{a} \dfrac{4}{b} \dfrac{4}{c} \dfrac{4}{d} }$ $= \boxed{\dfrac{4^5}{3^3}}$