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.

Let $f(a,b,c,d)=8(abcd+1)-(a+1)(b+1)(c+1)(d+1)$.
Hence, $f$ is a linear function of $a$, of $b$, of $c$ and of $d$. Thus,
$$\min_{\{a,b,c,d\}\subset[0,1]}f=\min_{\{a,b,c,d\}\subset\{0,1\}}f=f(1,1,1,1)=0$$
and since the minimum does not occur, we are done!
