Probability of at least two of $n$ independent events occurring subject to some conditions Given a set of independent Bernoulli random variables $\{x_1, \dots, x_n\}$, let $p = \sum_{0<i\leq n}\Pr[x_i = 1]$ and $X=\sum_{0<i\leq n} x_i$. We know that for any $i$, we have $\Pr[x_i = 1]\leq \frac{p}{2}$ and need to find a lower-bound for $\Pr[X>1]$ for any $p \in [0, 2]$. (The lower-bound should be a function of $p$.)
 A: This turned out stranger than I expected.
A good lower bound is
$$\min\left(\frac{p^2}{4}, 1-\frac{4+2p-p^2}{4e^{\,p/2}}, 1-\frac{1+p}{e^{\,p}} \right)$$
where the components apply in the regions [0,0.923], [0.923,1.547] and [1.547,2].
For $n=3$, I got an explicit solution from Mathematica with these minimizing probabilities and arguments:
\begin{align}
\frac{p^2}{4} \ \ \ \ \  &\text{ via }\{\frac{p}{2}, \frac{p}{2}, 0\}
\text{ if } p \in [0,1]\\
\frac{p^2(5-p)}{16} &\text{ via }\{\frac{p}{2}, \frac{p}{4}, \frac{p}{4}\}
\text{ if } p \in [1,\frac{9}{5}]\\
\frac{p^2(9-2p)}{27} &\text{ via }\{\frac{p}{3}, \frac{p}{3}, \frac{p}{3}\}
\text{ if } p \in [\frac{9}{5},2]\\
\end{align}
FullSimplify[Minimize[{q r + q s + r s - 2 q r s, p == q + r + s,
  0 <= q <= p/2, 0 <= r <= p/2, 0 <= s <= p/2}, {q, r, s}], Assumptions -> 0 < p < 2]

This and similar explicit results for $n=4$ suggest that for all $n$, the minimizing probabilities and arguments will be one of:
\begin{align}
\frac{p^2}{4} \ \ \ \ \  &\text{ via }\{\frac{p}{2}, \frac{p}{2}, 0, \ldots, 0\}
\\
1-\left(1+\frac{2p-p^2}{4-4q}\right)
\left(1-q\right)^{n-1}
&\text{ via }\{\frac{p}{2}, q,  \ldots, q\}
\\
1-\left(1+p-\frac{p}{n}\right)\left(1-\frac{p}{n}\right)^{n-1}
&\text{ via }\{\frac{p}{n}, \frac{p}{n},  \ldots, \frac{p}{n}\}
\end{align}
where $q=p/(2n-2)$.  The formula at the top comes from taking the limit of this as $n$ goes to infinity.
The graph shows the blue line for $n=3$, the orange line for $n=4$, and the green line for the limit.

A: As I see, for fixed $n$ and fixed $p$, the minimum of $P(X>1)$ can be expressed compactly as a recursive function of $n$ and $p$.
More precisely, let $G(n,p) = \text{min}( \mathbb{P}(X>1)$ under the given constraint and 
$$ F_n( y_1,y_2,y_3,...,y_n) = (1-y_1)(1-y_2)...(1-y_n)\left( 1+ \sum_{i=1}^n \frac{y_i}{1-y_i} \right) $$
As I observe, we can prove that :
Proposition 0 Forall $n>2$, 
$G(n,p) = \text{min}\left( G(n-1,p) , 1- F_n\left( q,q,...,q, \frac{p}{2}\right) , 1- F_n \left(\frac{p}{n},...,\frac{p}{n} \right) \right) $ where $q = \frac{p}{2n-2}$
However, I don't think these results are so obvious that we can state them without verification as in the answer of Matt.
