Let $\alpha\in[0,1]$ be a fixed constant, and let
$w,x\in[0,1]^n$ be two vectors such that $\sum_i w_i x_i=\alpha$.

Define $Y = \sum_i w_i X_i$, where $X_i \sim \operatorname{Bernoulli}(x_i)$, so it holds $\mathbb{E}[Y]=\alpha\leq 1$, and assume that the $X_i$'s are *independent* variables, .

Can we use some concentration inequality for $Y$ to obtain some bound of the form $Pr[Y>1] \leq c(\alpha)$ for some constant $c(\alpha)<1$ ?

**What I tried so far**

* The Chernoff-Hoeffding inequality gives me a bound of the form
$$
Pr[Y>1] \leq \exp\left(-\frac{2(1-\alpha)^2}{n}\right),
$$
but this is useless when $n\to\infty$.

* We can also use Bernstein inequality, as $\mathbb{V}[Y]=\sum_i w_i^2 x_i (1-x_i)\leq\sum_i w_i x_i=\alpha$, to obtain:
$$
Pr[Y>1] \leq \exp\left(-\frac{(1-\alpha)^2}{2(\alpha+\frac{1-\alpha}{3})}\right).
$$
This is better, but I don't really believe that this bound is tight. In particular, the above bound is trivial when $\alpha=1$, but intuitively I think it is possible to bound $Pr[Y<1]$ away from $1$.