Bonferroni-like inequality I am trying to prove this interesting tail bound for probabilities. Suppose that  $A_1,\dots,A_n$ are independent events. I am interested in showing that
$$
P(\bigcup_{I=1}^n A_i)\ge \sup_{m\ge 1}(1-e^{-m})\min\left\{1,\frac{1}{m}\sum_{i=1}^n P(A_i)\right\}.
$$
Not sure what can be done here, I tried using inclusion exclusion but that leads to no avail. Furthermore I have tried to just show directly this holds for just one of the $m$ but that doesnt really lead anywhere as well. Even just taking $m=1$, it's not quite clear why $(1-e^{-m})$ is the subadditivity gap.
 A: I suspect there's a better proof, and that such a proof would add some extra insight into the inequality.
Terms and preliminaries:
$$
\begin{align}
&a_i=P(A_i)\\
&Y=\sum a_i\\
\forall x>y>0, &\left( 1 - \tfrac yx\right)^x <e^{-y}
\end{align}
$$
The RHS is maximal at $m=Y$, where it is $1-e^{-Y}$. Proof: The derivatives rt $m$ for $m<Y$ and $m>Y$ are:$$
\begin{align}
m<Y:&\;\;e^{-m}>0\\
m>Y:&\;\;((m+1)e^{-m}-1)\tfrac Y{m^2}\le 0
\end{align}
$$
The latter follows from $e^m\ge m+1$. Since the function is continuous on $m>0$ and differentiable on $m\ne Y$, $m=Y$ is the maximum.
Note: this is why I think there must be a more interesting proof -- the $\sup$ is easily eliminated, which gives a simpler lower bound.
The LHS: Since he right side of the inequality depends only on $\sum P(A_i)$, consider what arrangement of probabilities with a fixed total minimizes the left side.
$$
\begin{align}
P(\bigcup_{i=1}^n A_i)&=1-\prod_{i=1}^n(1-a_i)\\
&\ge
 1-\max_{\sum x_i=Y\\
1\ge x_i\ge0}\prod_{i=1}^n(1-x_i) 
\end{align}
$$
It is pretty well known, and easy to show, that the maximum occurs when all terms are equal:$$
\max_{\sum x_i=Y\\
1\ge x_i\ge0}\prod_{i=1}^n(1-x_i) = \left( 1- \frac Yn \right)^n
$$
So it remains to show that:$$
1-\left( 1- \frac Yn \right)^n \ge 1-e^{-Y}
$$
This follows from the identity from the "preliminary" section.
A: Let $p = \sum_{i=1}^{n}P(A_i)$ and $m^* = \lfloor \sum_{i=1}^{n}P(A_i) \rfloor $. Note that, for $m \ge 1$,
\begin{align*}
g(m) &\overset{\text{def}}{=} (1-e^{-m}) \min\left(1, \frac{1}{m}p\right) \\
&= \min\left(1 - e^{-m}, \frac{1-e^{-m}}{m}p\right) \\
&= [m \le m^*](1 - e^{-m}) + [m > m^*]\frac{1-e^{-m}}{m}p
\end{align*}
where $[\cdot]$ denotes the Iverson bracket. Note that $g_1(m) = 1 - e^{-m}$ is increasing and $g_2(m) = \frac{1 - e^{-m}}{m}p$ is decreasing, so maximizing each region, we have
\begin{align*}
g(m) &\le [m \le m^*]f_1(m^*) + [m > m^*]f_2(m^*+1) \\
&\le [m \le m^*]f_1(p) + [m > m^*]f_2(p) \\
&= 1 - e^{-p}
\end{align*}
because $f_1(p) = f_2(p) = 1 - e^{-p}$ and $m^* \le p \le m^*+1$. Then, it becomes a straightforward to show $1 - e^{-p} \le 1 - \prod_{i=1}^{n}(1 - P(A_i)) = P(\cup_{i=1}^{n}A_i)$.
