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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

user635988
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  • Are there other constraints? As stated, the function has neither globally minimizing nor globally maximizing points. By taking $y,z:=\dfrac{1-x}{2}$, you can show that the function can be arbitrarily large both positively and negatively. If $x,y,z>0$ holds, then a simple AM-GM argument shows that the unique maximizing point is given by $x=y=z=\dfrac13$. – Batominovski Jun 15 '20 at 15:51
  • Isn't that give $x=0$? – user635988 Jun 15 '20 at 15:55
  • I dun include the case where $x,y,z$ equal 0 – user635988 Jun 15 '20 at 15:57
  • In my comment above, I made a typo. The last sentence should be "If $x,y,z>0$ holds, then a simple AM-GM argument shows that the unique *minimizing* point is given by $x=y=z=\dfrac13$." Using Shubhrajit Bhattacharya's answer below with a small $\epsilon<0$, you can show that your function has no globally minimizing points if you don't have the constraint $x,y,z>0$. – Batominovski Jun 15 '20 at 17:12

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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$}$

ShBh
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  • not related to the solve, but I'm wondering how do you justify the text inside [ ] on the right like this ? – zwim Jun 15 '20 at 17:17
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    Here is the TeX command \tag*{$\left[\text{since $\frac{2}{1-\varepsilon}>2$}\right]$} – ShBh Jun 15 '20 at 17:18