5
$\begingroup$

$x, y, z$ 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?

  • 1
    Is "$(1/x-1)$" supposed to be $\frac{1}{x-1}$, or $\frac{1}{x}-1$?2012-07-21
  • 0
    The latter one sir.2012-07-21
  • 0
    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 $\geq 8$.2012-07-21
  • 0
    Now the problem does make sense.2012-07-21
  • 0
    So can it be $16$ too? Please explain.2012-07-21
  • 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 $8$.2012-07-21
  • 0
    @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.2012-07-21
  • 0
    @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.2012-07-21
  • 0
    @Potato - my apologies - I failed to read the question. Didn't see the word "distinct". Sorry.2012-07-21

3 Answers 3

6

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 $z$.

2

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 $16.$

  • 0
    But $10$ is not in options..2012-07-21
  • 0
    Whose options? Where did you get this problem?2012-07-21
  • 0
    Sir, I edited the question, please have a look at the title.2012-07-21
2

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,

  • 0
    Yup I too thought so but the question didn't have the term "minimum" and thus the confusion. Thanks though.2012-07-21
  • 0
    We can't actually get $8$ because the problem stipulates that the numbers are distinct. It's very poorly written.2012-07-21