Find the maximum and minimum of $(1/x-1)(1/y-1)(1/z-1)$ if $x+y+z=1$ I get more confused when I try to solve this. For my first approach, I'm using normal AM-GM inequality:
\begin{equation}
  (\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)=\frac{(1-x)(1-y)(1-z)}{xyz} \\=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1
\end{equation}
By $x+y+z=1\geq \sqrt[3]{xyz}$ $\Rightarrow \frac{1}{27} \geq xyz$ we get
\begin{equation}
  (\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\geq 8
\end{equation}
So this is the minimum value. I do some research for how to find the maximum and come across an interesting Theorem " Lagrange Multipliers " than I'm using this to solve the problem. I'm not pretty good at Latex so I'm going for a shortcut. So we have $f(x,y,z)= (\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)$ and $g(x)=x+y+z-1$. After derivated I get:
\begin{equation}
 \frac{-1+z+y-yz}{x^2yz}=\frac{-1+x+z-xz}{xy^2z}=\frac{-1+x+y-xy}{xyz^2}=\lambda
\end{equation}
Solving this I get $x=y=1$, $z=-1$ and so on...this kind of solution give 0 for the answer. And $x=y=z=\frac{1}{3}$ which give 8. But that mean 8 is the maximum!? Please help

Actually the expression $$\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)$$ can be made arbitrarily large with the constraint $(x+y+z)=1$
Proof: Let $N>0$ be any arbitrary large real number. Choose $\varepsilon\in(0,1)$ such that $$\frac{1}{\varepsilon}>(N+1)$$ Then let $x=\varepsilon,y=z=\frac{1-\varepsilon}{2}$. Then clearly $$x,y,z>0,(x+y+z)=1\quad\text{and}\quad\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)=\left(\frac{1}{\varepsilon}-1\right)\left(\frac{2}{1-\varepsilon}-1\right)^2>\left(\frac{1}{\varepsilon}-1\right)(2-1)^2=\left(\frac{1}{\varepsilon}-1\right)>N$$ $$\tag*{$\left[\text{since $\frac{2}{1-\varepsilon}>2$}\right]$}$$
Hence the expression $$\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)$$
can be made arbitrarily large and hence attains no maximum in the set $$\mathcal{S}=\{(x,y,z):x,y,z>0;(x+y+z)=1\}$$ So $$\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)\in[8,\infty)\;\forall\;(x,y,z)\in\mathcal{S}$$
$\tag*{$\square$}$
