# An interesting puzzle

Here is a puzzle challenge for you: Suppose $X,Y$ are independent and identically distributed Random Variables. Show that $$P\{|X-Y|\le2\}\le 3P\{|X-Y|\le 1\}$$

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I think a rough sketch might be that in order to create any significant probability mass in $\vert X-Y \vert$ between 1 and 2, you have to also create mass between 0 and 1. So, $P{|X−Y|≤1}$ is no less than half the LHS? – Neil G Aug 11 '13 at 14:39