TITLE: To show that $P(|X-Y| \leq 2) \leq 3P(|X-Y| \leq 1)$
QUESTION [11 upvotes]: I found this question while browsing through "The Probabilistic Method", by Noga Elon.
Let X and Y be 2 independent and identically distributed real valued random variables. Prove that:
$$P(|X-Y| \leq 2) \leq 3P(|X-Y| \leq 1)$$
So I tried the following:
$$P\{|X-Y| \leq 2\} = P\{|X-Y| \leq 1\} + P\{X-Y \in (1,2]\cup[-2,-1)\}$$
$$= P\{|X-Y| \leq 1\} + P\{X-Y \in (1,2]\} + P\{X-Y \in [-2,-1)\}$$
$$= P\{|X-Y| \leq 1\} + 2P\{X-Y \in (1,2]\}$$
where the last step follows because $X-Y$ has a symmetric distribution. NB: A random variable Z has symmetric distribution if 
$$P(Z \leq z) = P(Z \geq -z) \quad \forall z \in \mathbb{R}$$
Thus the problem boils down to showing
$$P\{X-Y \in (1,2]\} \leq P(|X-Y| \leq 1)$$
and I would be done. Unfortunately, I don't know how to proceed from here. I appreciate any help, hints, useful comments etc. I receive.

REPLY [4 votes]: A proof is given in "The 123 Theorem and its extensions" by Noga Alon, Raphael Yuster. (See also this question.)