<s>[INST] 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.
 [/INST] 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 .
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