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Any help? I reached this stage:

$$[(x-z)^2]\cdot 17^{-1} + (x-z)^{-1}$$

Is this even correct? If so, where do I go from there?

Thank you.

PS: This question couldn't be solved by my maths teacher, so I hope to receive any help from here.

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  • $\begingroup$ I doubt it! As none of the numbers equal one another, and the differences can be multiplied to generate a prime number (17), it can be assumed that not all of the numbers are integers. I hope what I said makes sense.. $\endgroup$ – StaticCrazee Sep 30 '15 at 20:24
  • $\begingroup$ Unless there's a hidden assumption that $x,y,z$ are integers, this has nothing to do with LCMs or factorization at all. (And if they are supposed to be integers, then the title refers to an LCM but the statement refers to a product; they are not the same thing.) $\endgroup$ – Greg Martin Sep 30 '15 at 20:28
  • $\begingroup$ Alright. I may have got the wording wrong, but the product also gives the LCM, be it useful or not. I'm just stating it, might be helpful in providing a solution? $\endgroup$ – StaticCrazee Sep 30 '15 at 20:34
  • $\begingroup$ Assuming you're talking about integers in the first place: no, the LCM and the product are not the same thing. What is the LCM of 4 and 6? The product? $\endgroup$ – Greg Martin Oct 1 '15 at 6:49
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Let $m = x-y, n = y-z$, and then $m+n = x-z$. We then attempt to minimize

\begin{align} \frac{1}{m}+\frac{1}{n}+\frac{1}{m+n} & = \frac{mn+m^2+mn+n^2+mn}{mn(m+n)} \\ & = \frac{(m+n)^2+mn}{17} \\ & = \frac{(m+n)^2+\frac{17}{m+n}}{17} \end{align}

Now let $u = m+n = x-z$, and we are attempting to minimize

$$ \frac{u^2+\frac{17}{u}}{17} = \frac{u^2}{17}+\frac{1}{u} $$

This is where you ended up, I believe. Differentiation helps best here; the minimum value is achieved when

$$ \frac{d}{du} \left( \frac{u^2}{17}+\frac{1}{u} \right) = \frac{2u}{17}-\frac{1}{u^2} = 0 $$

or

$$ u^3 = \frac{17}{2} $$

Substituting back into our previous expression yields a minimum value of

$$ \frac{u^3+17}{17u} = \frac{\frac{17}{2}+17}{17u} = \frac{\frac{51}{2}}{17u} = \frac{3}{2u} = \sqrt[3]{\frac{27}{68}} \doteq 0.735 $$

There might well be a simpler, more elegant solution to this problem, but this is reasonably quick.

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  • $\begingroup$ If you don't like the tiny $3$ for the cube root, you can see suggestions here on tex.stackexchange and on the linked question. I was writing up the same solution when yours posted. +1 $\endgroup$ – Ross Millikan Sep 30 '15 at 20:41

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