Independent Bernoulli random variables My professor left us to solve this problem:
Let $\xi_1, \xi_2,...,\xi_n$ be independent Bernoulli random variables defined on a probability space $(\Omega,P(\Omega),\mathbb{P})$ such that:
$\mathbb{P}(\xi_i=0) = 1 -\lambda_i\Delta$, $\mathbb{P}(\xi_i=1) = \lambda_i\Delta$, where $\lambda_1, \lambda_2,...,\lambda_n$ are positive, and $0<\Delta<\frac{1}{max\{\lambda_1,\lambda_2,...\lambda_n\}}$.
Prove that $\mathbb{P}(\xi_1+\xi_2+...+\xi_n = 1) \geq \Delta\sum_{i=1}^{n}\lambda_i - \Delta^2(\sum_{i=1}^n\lambda_i)^2$.
I can prove that $\mathbb{P}(\xi_1+\xi_2+...+\xi_n = 1) \leq \Delta\sum_{i=1}^{n}\lambda_i$ since $\mathbb{P}(\xi_1+\xi_2+...+\xi_n = 1) = \sum_{i=1}^n\lambda_i\Delta(\Pi_{j=1\\j\neq i}^n1-\lambda_j)^2 \leq \Delta\sum_{i=1}^n\lambda_i$ but I don't know how to prove another inequality.

You can rewrite
$$\Delta\sum_{i=1}^{n}\lambda_i - \Delta^2(\sum_{i=1}^n\lambda_i)^2 = \sum_{i=1}^{n} \left[\Delta \lambda_i \cdot \left( 1 - \sum_{i=1}^n \Delta\lambda_i\right) \right] $$
And compare with the exact expression
$$\sum_{i=1}^{n} \left[\Delta \lambda_i \cdot \prod_{j\neq i}\left( 1 -  \Delta\lambda_j\right) \right] $$
This product $ \prod_{j\neq i}\left( 1 -  \Delta\lambda_j\right)$ expressing the probability that the other $\xi_j = 0$ when $\xi_i =1$ is simplified with a sum that underestimates this value.
This is similar to the difference between the Bonferroni correction and the Šidák correction. You can proof this with Boole's inequality.
