In the context of Hurwitz groups and manifolds, one comes by what wikipedia defines as a "remarkable" fact, that $1-\dfrac1 a -\dfrac 1 b - \dfrac1 c > 0$ has a minimal value of $1/42$ if $a,b,c \in \mathbb{Z}$ and $a<b<c$ (I'm not sure this last part is needed, but let's consider it anyway).

I was wondering how one would go about solving a problem like this. I am totally at loss when I see problems "over the integers": I never actually learned any techniques to approach such problems. So the question is: is there a suitable general method to minimize expressions over the integers? I of course feel that there is not standard way, but I'm interested in some examples as to how to proceed, maybe taking this expression as a base for the examples.

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    $\begingroup$ I am not sure you have quite specified the problem correctly. We have $\frac12+\frac13+\frac17-1=-\frac1{42} \lt0$. Note that for the sum of three reciprocals to be greater than $1$ we must have one fraction greater than $\frac13$, so we need $a<3$ so $a=2$. Similarly we need $b<4$ so $b=3$, and then $c<6$ so $c=4$ or $c=5$ - so there are only two possibilities to test. $\endgroup$ – Mark Bennet May 27 '13 at 8:32
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    $\begingroup$ I think the correct statement of the problem is to minimize $1-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}>0$. In your statement a minimizer doesn't exist since the infinum is easily seen to be $0$. $\endgroup$ – Julian May 27 '13 at 8:35
  • $\begingroup$ Thanks for pointing that out! I edited it. $\endgroup$ – Andy May 27 '13 at 8:49

For this problem you just need some intelligent case analysis. Increasing the value of $a, b$ or $c$ makes the sum of reciprocals smaller and thus increases the expression on the right hand side. So you need the smallest values you can get.

$a=1$ leaves the left hand side negative, and the best we could do with $a\ge 3$ would be to set $a=3, b=4, c=5$ giving us $\frac {13}{60}$. So we need to test $a=2$ to see if we can do better than this.

The expression then becomes $\frac12-\frac1b-\frac1c$ with $2\lt b \lt c$

If $b\ge 4$ the best we can do is $b=4, c=5$, giving us a value of $\frac 1{20}$.

So we are left testing $b=3$, when our expression becomes $\frac16-\frac1c$. The lowest value of $c$ for which this is positive is $c=7$ and this gives the minimum value $\frac1{42}$.

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  • $\begingroup$ Note: The trial and error/case analysis approach works here because the search is very easily constrained. $\endgroup$ – Mark Bennet May 27 '13 at 9:23

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