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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?

J. M. ain't a mathematician
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Hyperbola
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    Is "$(1/x-1)$" supposed to be $\frac{1}{x-1}$, or $\frac{1}{x}-1$? – Arturo Magidin Jul 21 '12 at 04:13
  • The latter one sir. – Hyperbola Jul 21 '12 at 04:14
  • 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 $\geq 8$. – Sangchul Lee Jul 21 '12 at 04:26
  • Now the problem does make sense. – Sangchul Lee Jul 21 '12 at 04:27
  • So can it be $16$ too? Please explain. – Hyperbola Jul 21 '12 at 04:27
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    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 05:19
  • @Potato Indeed you CAN make the expression arbitrarily large. What you can NOT do is make the expression arbitrarily SMALL. In fact, you can't make it any smaller than 8, with the given constraints. I'm not sure why you assert that there is no minimum possible value. –  Jul 21 '12 at 08:45
  • @David Wallace There is no minimal element in the set of attainable values, because 8 is not attainable, but every number greater than 8 is. – Potato Jul 21 '12 at 17:48
  • @Potato - my apologies - I failed to read the question. Didn't see the word "distinct". Sorry. –  Jul 21 '12 at 20:49

3 Answers3

7

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 $z$.

André Nicolas
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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 $16.$

Will Jagy
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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,

Patrick Da Silva
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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 05:01
  • 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 05:33