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.
 A: 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 $0 < Q = abc < 1.$
A: 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
$$0<a,b,c,d<1$$
A: 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!
A: Given that $0<a<1$, $0<b<1$, $0<c<1$, $0<d<1$
$\therefore$ $-1<a-1<0$ and $-1<b-1<0$
Now
\begin{align*}
&(a-1)(b-1)>0 \\
\Rightarrow& ab - (a+b) +1 >0 \\
\Rightarrow& (ab+1) > a+b \\
\Rightarrow& (ab+1)+(ab+1) > (ab+1)+(a+b) \\
\Rightarrow& 2(ab+1) > (ab+1)+(a+b) \\
\Rightarrow& 2(ab+1) > (a+1)(b+1) \tag{I}
\end{align*}
Similarly
$$2(cd+1) > (c+1)(d+1)$$
$\therefore$ $0<a<1$ and $0<b<1$ $\Rightarrow$ $0<ab<1$.
Similarly $0<cd<1$.
Now,
\begin{align*}
4(ab+1)(cd+1) &> (a+1)(b+1)(c+1)(d+1) \\
\Rightarrow (a+1)(b+1)(c+1)(d+1) &< 4(ab+1)(cd+1) \tag{II}
\end{align*}
As $(I)$
\begin{align*}
2(abcd+1) &> (ab+1)(cd+1) \\
\Rightarrow (ab+1)(cd+1) &< 2(abcd+1) \tag{III}
\end{align*}
Putting these value in $(II)$,we get
\begin{align*}
(a+1)(b+1)(c+1)(d+1) &< 4\left\{2(abcd+1)\right\} \\
(a+1)(b+1)(c+1)(d+1) &< 8(abcd+1)
\end{align*}
answered by SAROJ GHOSH .
