How to prove:
$$|x+y| \cdot \mathbb{I}_{\{|x+y| \ge 2a \}} \le 2\big( |x| \cdot \mathbb{I}_{\{ |x| \ge a\}} + |y| \cdot \mathbb{I}_{\{|y| \ge a\}} \big) $$
My attempt,
I know that $ \{ |x + y| \ge 2a\} \subseteq \{|x| \ge a\} \cup\{|y| \ge a\} $. As a result, we get: \begin{align*} & |x+y| \cdot \mathbb{I}_{\{ |x+y| \ge 2a\}} \le (|x+y|) \cdot \mathbb{I}_{\{|x| \ge a\} \cup\{|y| \ge a\}} \\ &\hspace{3.35cm} \le |x| \cdot \mathbb{I}_{\{|x| \ge a\} \cup\{|y| \ge a\}} + |y| \mathbb{I}_{\{|x| \ge a\} \cup\{|y| \ge a\}} \end{align*}
Now I am lost.. I don't understand or get how to get rid of the union of events and evaluate them separately. I've tried a couple of different ideas - but they are all futile.
Any suggestions are appreciated - thanks a lot.
If $|x+y|<2a$, the LHS is zero and the inequality is trivial.
Proceed assuming $|x+y|\geq2a$. It follows that $$ a\leq\frac{1}{2}\left|x+y\right|\leq\frac{1}{2}\left(\left|x\right|+\left|y\right|\right)\leq\max\left\{ \left|x\right|,\left|y\right|\right\} . $$ W.l.o.g., assume $|x|\geq|y|$ (if this is not true, relabel $x$ and $y$). By the above, $|x|\geq a$ and hence \begin{align*} \left|x+y\right|\mathbb{I}_{\left\{ \left|x+y\right|\geq2a\right\} } & =\left|x+y\right| & \text{(}\left|x+y\right|\geq2a\text{)}\\ & \leq\left|x\right|+\left|y\right| & \text{(triangle ineq.)}\\ & \leq2\left|x\right| & \text{(}\left|x\right|\geq\left|y\right|\text{)}\\ & =2\left|x\right|\mathbb{I}_{\left\{ \left|x\right|\geq a\right\} } & \text{(}\left|x\right|\geq a\text{)}\\ & \leq2\left(\left|x\right|\mathbb{I}_{\left\{ \left|x\right|\geq a\right\} }+\left|y\right|\mathbb{I}_{\left\{ \left|y\right|\geq a\right\} }\right) & \text{(}\left|y\right|\mathbb{I}_{\left\{ \left|y\right|\geq a\right\} }\geq0\text{)} \end{align*}