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$x, y, z$ 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?

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1  
Is "$(1/x-1)$" supposed to be $\frac{1}{x-1}$, or $\frac{1}{x}-1$? –  Arturo Magidin Jul 21 '12 at 4:13
    
The latter one sir. –  Hyperbola Jul 21 '12 at 4:14
    
It's minimum is clearly $8$ (achieved when $x = y= z = \frac{1}{3}$), as you have shown. But that's it. It can attain any value $\geq 8$. –  sos440 Jul 21 '12 at 4:26
    
Now the problem does make sense. –  sos440 Jul 21 '12 at 4:27
2  
There is no minimum possible value. The minimum $8$ is only achieved when $x=y=z$, which is ruled out (distinct positive reals). You can, however, get arbitrarily close to $8$. –  Potato Jul 21 '12 at 5:19

3 Answers 3

up vote 6 down vote accepted

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 $z$.

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Try $x = \frac{1}{2}, \; y = \frac{1}{3}, \; z = \frac{1}{6}.$ Your product is $$ (2-1)(3-1)(6-1) = 1 \cdot 2 \cdot 5 = 10. $$

Next, try Try $x = \frac{4}{7}, \; y = \frac{2}{7}, \; z = \frac{1}{7}.$ Your product is $$ (\frac{7}{4}-1)(\frac{7}{2}-1)(7-1) = \frac{3}{4} \cdot \frac{5}{2} \cdot 6 = \frac{45}{4}. $$

Your question has no fixed answer.

Take $x = \frac{10}{15}, \; y = \frac{3}{15}, \; z = \frac{2}{15}.$ Your product is $13.$

Take $x = \frac{6}{9}, \; y = \frac{2}{9}, \; z = \frac{1}{9}.$ Your product is $14.$

Take $x = \frac{15}{20}, \; y = \frac{3}{20}, \; z = \frac{2}{20}.$ Your product is $17.$

Take $x = \frac{14}{21}, \; y = \frac{6}{21}, \; z = \frac{1}{21}.$ Your product is $25.$

Take $x = \frac{165}{252}, \; y = \frac{77}{252}, \; z = \frac{10}{252}.$ Your product is $29.$

Take $x = \frac{21}{28}, \; y = \frac{6}{28}, \; z = \frac{1}{28}.$ Your product is $33.$

Take $x = \frac{65}{78}, \; y = \frac{10}{78}, \; z = \frac{3}{78}.$ Your product is $34.$

Take $x = \frac{35}{50}, \; y = \frac{14}{50}, \; z = \frac{1}{50}.$ Your product is $54.$

Take $x = \frac{85}{102}, \; y = \frac{15}{102}, \; z = \frac{2}{102}.$ Your product is $58.$

Take $x = \frac{170}{294}, \; y = \frac{119}{294}, \; z = \frac{5}{294}.$ Your product is $62.$

Take $x = \frac{270}{297}, \; y = \frac{22}{297}, \; z = \frac{5}{297}.$ Your product is $73.$

Take $x = \frac{77}{99}, \; y = \frac{21}{99}, \; z = \frac{1}{99}.$ Your product is $104.$

Take $x = \frac{247}{364}, \; y = \frac{114}{364}, \; z = \frac{3}{364}.$ Your product is $125.$

Take $x = \frac{90}{126}, \; y = \frac{35}{126}, \; z = \frac{1}{126}.$ Your product is $130.$

It seems likely that the target 16 requires at least two of $x,y,z$ to be irrational. Certainly you can fix, for example, $x = 1/2$ and solve for $y,z.$ So, take $$x = \frac{1}{2}, \; \; y = \frac{1}{4} + \frac{1}{4} \sqrt{\frac{7}{15}}, \; \; z = \frac{1}{4} - \frac{1}{4} \sqrt{\frac{7}{15}}.$$ Your product is $16.$

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But $10$ is not in options.. –  Hyperbola Jul 21 '12 at 4:15
    
Whose options? Where did you get this problem? –  Will Jagy Jul 21 '12 at 4:23
    
Sir, I edited the question, please have a look at the title. –  Hyperbola Jul 21 '12 at 4:24

If you look at the statement of the AM-HM inequality correctly, you'd see that $$ \frac 1x + \frac 1y + \frac 1z \ge 9 $$ and equality only happens when $x=y=z$. Therefore we can assume equality happens to find the minimum value, but $x+y+z = 1$ implies $x=y=z=1/3$. Therefore the minimum is $8$ and is attained uniquely at the point $(1/3, 1/3, 1/3)$. Does that clarify the doubts you had?

Hope that helps,

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Yup I too thought so but the question didn't have the term "minimum" and thus the confusion. Thanks though. –  Hyperbola Jul 21 '12 at 5:01
    
We can't actually get $8$ because the problem stipulates that the numbers are distinct. It's very poorly written. –  Potato Jul 21 '12 at 5:33

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