Proof of a probability inequality claim I'm reading a paper which makes an assertion of this form with no proof, where $\mu_S$ is a sample mean random variable, $\mu_T$ is a true or population mean r.v, and $\varepsilon$ is a constant:

For every real number $k \in (0,1)$
$$\Pr[|\mu_{S2} - \mu_{S1}| \geq \varepsilon] \leq \Pr[|\mu_{S2} - \mu_{T}| \geq k \varepsilon] + \Pr[|\mu_{T} - \mu_{S1}| \geq (1-k)\varepsilon]$$

This isn't really obvious to me. How do I prove this, or at least satisfy myself that this is true?
For anyone interested in the source, this is from Bifet & Gavalda 2006. Appendix A Theorem 1.2.
 A: I find it helpful to think about things like this by "translating" events into logical statements, and set operations into logic operation.
Let $A = \{\mu_{S2} - \mu_{S1}| \geq \epsilon\}$, $B = \{|\mu_{S2} - \mu_{T}| \geq k\epsilon\}$ and $C = \{|\mu_{T} - \mu_{S1}| \geq (1-k)\epsilon\}$.  We want to show $A \subseteq B \cup C$, for then we will have $P(A) \le P(B \cup C) \le P(B) \cup P(C)$ by union bound.  The set relation $A \subseteq B \cup C$ is saying "if $A$ then $B$ or $C$".  Ok, well, we know how to prove statements of that form.  In this case it might be easiest to prove the contrapositive "if not $B$ and not $C$ then not $A$" (or in set notation $B^c \cap C^c \subseteq A^c$).
So suppose not $B$ and not $C$, i.e. $|\mu_{S2} - \mu_{T}| < k\epsilon$ and $|\mu_{T} - \mu_{S1}| < (1-k)\epsilon$. (More precisely, suppose $\omega \in \Omega$ is such that these two statements hold for $\omega$).  Our goal is to prove not $A$, i.e. to upper bound $|\mu_{S2} - \mu_{S1}|$, and we can clearly see to use the triangle inequality:
$$|\mu_{S2} - \mu_{S1}| \le |\mu_{S2} - \mu_{T}| + |\mu_{T} - \mu_{S1}| < k \epsilon + (1-k)\epsilon = \epsilon$$
which is "not $A$".
