How to apply triangle inequality in a probability statement? $$\begin{align} P\left(\left| \bar{Y}+X_n-\mu\right| \ge \delta \right) &\le P\left( \left|\bar{Y}-\mu\right| \ge \frac{\delta}{2} \cup \left|X_n\right| \ge \frac{\delta}{2} \right)  \tag{1} \\
& \le P\left( \left|\bar{Y}-\mu\right| \ge \frac{\delta}{2} \right) + \left( \left|X_n\right| \ge \frac{\delta}{2} \right) \tag{2}
\end{align}$$
I don't know how to get the right hand side of (1).
I do know that by the triangle inequality: $$\left| \bar{Y}+X_n-\mu\right|\le \left| \bar{Y}-\mu\right| + \left|X_n\right|$$
But am unsure how that translate in terms of probability. This is the first time I am seeing such an application of the triangle inequality and my text simply states it without further explanation. Please advice.
Also, would such an inequality hold if the relation in $P(\dots)$ was $\le$ instead?
 A: Hint: prove that
$$
\{\omega\in\Omega;|\bar{Y}(\omega)+X_n(\omega)-\mu| \ge \delta \} 
\subseteq  
\{\omega\in\Omega;|\bar{Y}(\omega)-\mu| \ge \frac{\delta}{2}\} 
\cup 
\{\omega\in\Omega;|X_n(\omega)| \ge \frac{\delta}{2}\}   
$$
Solution:
Let $\omega\in\{\omega\in\Omega;|\bar{Y}(\omega)+X_n(\omega)-\mu| \ge \delta \} 
$, then if $\omega\notin\{\omega\in\Omega;|X_n(\omega)| \ge \frac{\delta}{2}\}$
one has
$$
|\bar{Y}(\omega)+X_n(\omega)-\mu| \ge \delta ,\quad |X_n(\omega)| < \frac{\delta}{2}
$$
So using triangle inequality one has
$$
|\bar{Y}(\omega)-\mu| \ge |\bar{Y}(\omega)+X_n(\omega)-\mu|- |X_n(\omega)| \ge \delta- \frac{\delta}{2}=\frac{\delta}{2}.
$$
Thus
$$\omega\in\{\omega\in\Omega;|\bar{Y}(\omega)-\mu|  \ge \frac{\delta}{2}\}$$   
A: The condition $|\bar{Y}+X_n-\mu | \geq \delta $ implies by the triangle inegality that $|\bar{Y}-\mu | \geq \delta /2$ or $|X_n | \geq \delta /2$. 
A: I can't comment yet, but I came across this question, and wanted to add that @mauna's final question would not hold. The reason this all works is because triangle ineq gives us that $$|x|+|y| \geq |x+y| \geq \delta$$ and if the latter inequality was switched, we wouldn't have anything to work with... Concretely, if $x=-3,y=4$, and $\delta = 1.1$, we would have that $|-3+4|=1\leq 1.1$ but dividing the absolute value of neither would be true. However, if we knew that $\delta \leq |x+y|$, we could go in the other way...
