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.
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.
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?