Let $X_1, \cdots, X_k$ be identically distributed. Does it hold that $P[\sum X_i > t] \leq 1-F_{X_1}(t/k)$ Let $X_1, \cdots, X_k$ be identically distributed, but not necessarily independent. I want to bound $P[\sum X_i > t]$. Intuitively speaking, the worst case should be if $X_1 = X_2 = \cdots = X_k$. If this were true, I would get that $P[\sum X_i > t] \leq 1-F_{X_1}(t/k)$. Is this so?
 A: As what you have mentioned, we just need to write them down a little bit more formally.
Note that
$$ X_i \leq \frac {t} {k} ~ \forall i \in \{1, 2, \ldots, k\} \Rightarrow \sum_{i=1}^k X_i \leq t $$
by summing the inequalities. The contra-positive statement will be
$$ \sum_{i=1}^k X_i > t \Rightarrow X_i > \frac {t} {k} ~ \exists i \in \{1, 2, \ldots, k\} $$
Express this in terms of the terminology of events, it becomes
$$ \left\{\sum_{i=1}^k X_i > t \right\} \subseteq \bigcup_{i=1}^k \left\{X_i > \frac {t} {k}\right\}$$
Finally to compute the probability,
$$ \begin{align} \Pr\left\{\sum_{i=1}^k X_i > t \right\} 
&\leq \Pr\left\{\bigcup_{i=1}^k \left\{X_i > \frac {t} {k}\right\}\right\} \\
&\leq \sum_{i=1}^k \Pr\left\{X_i > \frac {t} {k}\right\}  \\
&= \sum_{i=1}^k \left[1 - F_{X_i}\left(\frac {t} {k}\right)\right] \\
&= k\left[1 - F_{X_1}\left(\frac {t} {k}\right)\right]
\end{align}$$
where the second inequality is the Boole's inequality, and the last equality follows from the fact that $X_i$ are identically distributed.
It looks like you will need the constant $k$.
Edit: Counter example:
Suppose $k = 2$, $X_1 = 1 - X_2 \sim \text{Uniform}(0, 1)$, $t \in (0, 1)$.
Then $X_1 + X_2 = 1$ almost surely and thus
$$ \Pr\{X_1 + X_2 > t\} = \Pr\{1 > t\} = 1 $$
$$ 1 - F_{X_1}\left(\frac {t} {2}\right) = 1 - \frac {t} {2} < 1 = \Pr\{X_1 + X_2 > t\}$$
