Inequality Of Four Variables Suppose $a,b,c,d$ are real numbers greater than $1$. Prove that
$$abc+abd+acd+bcd-3abcd<1.$$

The inequality holds because
\begin{align*}
&4\times\left[\,3abcd-(abc+abd+acd+bcd)+1\,\right]\\
\\
=&2(a-1)(b-1)(c-1)(d-1)\\
\\
&+(a+1)(b+1)(c-1)(d-1)+(a+1)(b-1)(c+1)(d-1)\\
&+(a+1)(b-1)(c-1)(d+1)+(a-1)(b+1)(c+1)(d-1)\\
&+(a-1)(b+1)(c-1)(d+1)+(a-1)(b-1)(c+1)(d+1)\\
\\
&+(a+1)(b-1)(c-1)(d-1)\\
&+(a-1)(b+1)(c-1)(d-1)\\
&+(a-1)(b-1)(c+1)(d-1)\\
&+(a-1)(b-1)(c-1)(d+1).
\end{align*}
