I don't know where to post this question, so I am posting it here.

Take any n points in a plane and let D and d be the greatest and least distances determined by points of this set. Prove that 2D > sqrt(3) [sqrt(n)-1]d.

If it is easy, give hint.

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- Jun 5th 2006, 07:00 AMmalaygoelDistance between points
I don't know where to post this question, so I am posting it here.

Take any n points in a plane and let D and d be the greatest and least distances determined by points of this set. Prove that 2D > sqrt(3) [sqrt(n)-1]d.

If it is easy, give hint. - Jun 6th 2006, 12:06 PMThePerfectHacker
malaygoel.

Please do not make begging posts, everyone can see your question.

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I was not able to find a way, but I was thinking maybe you can use induction. Assume it works for $\displaystyle n$ then it must work for $\displaystyle n+1$. The first case with $\displaystyle n=3$ is clearly true because it leads to,

$\displaystyle 2D\geq \sqrt{3}(\sqrt{3}-1)d$

Which is clearly true because,

$\displaystyle D\geq d$

$\displaystyle 2\geq \sqrt{3}(\sqrt{3}-1)$

But for $\displaystyle n=4$ I was having difficulty.

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How do you know that this inequality is true?

Is it a known theorem?