Inequality for positive real numbers less than $1$: $8(abcd+1)>(a+1)(b+1)(c+1)(d+1)$ If $a,b,c,d$ are positive real numbers, each less than 1, prove that the following inequality holds:
$$8(abcd+1)>(a+1)(b+1)(c+1)(d+1).$$ 
I tried using $\text{AM} > \text{GM}$, but I could not prove it.

your inequality is equivalent to
$$7\,abcd-abc-abd-acd-bcd-ab-ac-ad-bc-bd-cd-a-b-c-d+7>0$$ combining usefull sums we obtain
$$(1-a)(1-bcd)+(1-b)(1-acd)+(1-c)(1-abd)+(1-d)(1-abc)+(1-ad)(1-bc)+(1-bd)(1-ac)+(1-cd)(1-ab)>0$$ since we have
$$0<a,b,c,d<1$$
