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.

First, note that $2^0(a+1) \ge a + 1$.
Then we prove by induction that $2^{k-1}(x_1\dots x_k+1) > \prod_{i=1}^k(x_i + 1)$. For the ease of notation, there I will focus on the case $k = 4$.
$$(d+1)(a+1)(b+1)(c+1)\le 4(d+1)(abc+1)< 4(abcd+abc+1+d)< 8(abcd+1)$$
The last inequality is provided by
$$abc+d< abcd+1 \Leftarrow Q+d<Qd+1 \Leftarrow (Q-1)(d-1)>0,$$
where $0 < Q = abc < 1.$
