<s>[INST] if $(1-a)(1-b)(1-c)(1-d) = \frac{9}{16}$ then minimum integer value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = ?$ Given  $a,b,c,d > 0$, how do we find the minimum integer value of $n=\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$ such that   $(1-a)(1-b)(1-c)(1-d) = \frac{9}{16}$.
 [/INST] If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq 1$, then $a,b,c,d>1$ and
$$\begin{align}
0\leq abcd-bcd-cda-dab-abc=&(1-a)(1-b)(1-c)(1-d)-1+(a+b+c+d)
\\&-(ab+ac+ad+bc+bd+cd)
\\
=&\frac{9}{16}-1-a(b-1)-b(c-1)-c(d-1)-d(a-1)
\\
&-ac-bd
\\
<&-\frac{7}{16}\,,
\end{align}
$$
which is a contradiction.  Hence, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}>1$.  Now, $$(a,b,c,d)=\left(\frac{161-\sqrt{8257}}{64},2,3,\frac{161+\sqrt{8257}}{64}\right)$$
is a solution to $(a-1)(b-1)(c-1)(d-1)=\frac{9}{16}$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=2$.  Ergo, the minimum possible integer value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ is $2$.
I think it is a nice question to ask whether there exists a solution $(a,b,c,d)$ with $a,b,c,d>0$ to $(a-1)(b-1)(c-1)(d-1)=\frac{9}{16}$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=t$ for any $t>1$.  I conjecture that the answer is positive.
</s>