The minimum value of $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)$ if $x+y+z=1$ $x, y, z$ are three distinct positive reals such that $x+y+z=1$, then the minimum possible  value of $(\frac{1}{x}-1) (\frac{1}{y}-1) (\frac{1}{z}-1)$ is ?
The options are: $1,4,8$ or $16$ 
Approach: $$\begin{align*}
\left(\frac{1}{x} -1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)&=\frac{(1-x)(1-y)(1-z)}{xyz}\\
&=\frac{1-(x+y+z)+(xy+yz+zx)-xyz}{xyz}\\
&=\frac{1-1+(xy+yz+zx)-xyz}{xyz}\\
&=\frac{xy+yz+zx}{xyz} - 1
\end{align*}$$
Now by applying $AM≥HM$, I got the least value of $(xy+yz+zx)/xyz$ as $9$, so I got final answer as $8$. Is it correct?

If we put no constraint on $x$, $y$, and $z$ apart from $x$, $y$, $z$ positive and $x+y+z=1$, then indeed your calculation, and the one by Patrick Da Silva, show that the minimum value is $8$, attained at $x=y=z=\frac{1}{3}$. 
However, the problem specifies that $x$, $y$ and $z$ are distinct real numbers. If we take that constraint into account, there is no minimum. We can get arbitrarily close to $8$ (but above $8$) by choosing $x$, $y$, and $z$ distinct and close to $\frac{1}{3}$, but we cannot attain $8$ with distinct $x$, $y$ and $z$. 
