absolute value of sum vs sum of absolute values I know that if $w$ satisfies $\lvert f_1(w)+f_2(w)\rvert>b$ then $\lvert f_1(w)\rvert>b/2$ or $\lvert f_2(w)\rvert>b/2$, given $b>0$ and $f_1,f_2$ two functions. This can be viewed geometrically.
If $f_1, f_2$ are measurable functions, this implies the well known result that $$P(\lvert f_1+f_2\rvert>b)\leq P(\lvert f_1\rvert>b/2)+P(\lvert f_2\rvert>b/2). \quad (1)$$
Question. Is it true that $$P(\lvert \sum_{k=1}^n f_k\rvert>b)\leq \sum_{k=1}^n P(\lvert f_k\rvert>b/n)? \quad (2)$$
How to prove it?
Comment. For $n=3$, I could use $(1)$ repeatedly, as
\begin{align}
P(\lvert f_1+f_2+f_3\rvert>b)&\leq P(\lvert f_1+f_2\rvert>b/2)+P(\lvert f_3\rvert>b/2)\\
&\leq P(\lvert f_1>b/4)+P(\lvert f_2\rvert>b/4)+P(\lvert f_3\rvert>b/2).
\end{align}
Clearly, this inequality is different from $(2)$. This example suggests me something like
$$P(\lvert \sum_{k=1}^n f_k\rvert>b)\leq \sum_{k=1}^n P(\lvert f_k\rvert>b/2^{\lfloor (n-1)/2\rfloor +1}). \quad (4)$$
But $(2)$ is a better bound than $(4)$.

Suppose $x \in \{\sum_{k\leq n} |f_k|>b\}$ but $x \notin \cup_{k\leq n}\{|f_k|>b/n\}$. Then $|f_k(x)|\leq b/n,\,\forall k\leq n$. This would imply $\sum_{k\leq n}|f_k(x)|\leq b$, which is a contradiction. Therefore $\{\sum_{k\leq n} |f_k|>b\}\subseteq \cup_{k\leq n}\{|f_k|>b/n\}$.
