Let $X\_1,\dots,X\_n$ be independent Bernoulli variables with probability $p<\frac{1}{2}$ (even $p\le\frac{1}{3}$ if needed). Let $\alpha\_1,\ldots,\alpha\_n$ be non-negative real numbers such that $\sum\limits\_{i=1}^n \alpha\_i = 1$, and let $Y = \sum\limits\_{i=1}^n \alpha\_i X\_i$.

Prove (or disprove): $\Pr(Y\ge p)\ge p$.

I can prove this for all $p=\frac{1}{k}$, $k\ge3$, but intuitively it seems to me that it should be true for all $p<\frac{1}{2}$, and I could not find a proof for that.

---

EDIT: my (or rather, a friend's) proof for $p = \frac{1}{k}$:

For each Bernoulli variable define a random variable $Z\_i$ that takes values from a set of formal objects $\{b\_1,\ldots b\_k\}$ uniformly, and define a function $f$ on these formal objects as $f(b\_1) = 1$, $\forall\_{i>1}f(b\_i)=0$.

Clearly, $f(Z\_i)$ is a Bernoulli variable with probability $p$, but the sample space for a convex combination of $f(Z\_1),\ldots f(Z\_n)$ can be now viewed as consisting of $k^n$ states of the form $(b\_{i\_1},\ldots,b\_{i\_n})$ from which one is selected uniformly.

Group these states into sets of $k$ states of cyclic shifts of the indices (that is, $(b\_{i\_1},\ldots,b\_{i\_n})$ and $(b\_{j\_1},\ldots,b\_{j\_n})$ are in the same set iff for all for all $1\le s \le n$, $b\_{j\_s} = b\_{i\_s} + c (\mod k)$ for some $s$-independent integer $c$).

The sum of the values of a convex combination of such sets under $f$ is exactly $1$, since each component is $b\_0$ in exactly one member of the set, so in each set there exists at least one state whose value is greater than or equal to $\frac{1}{k}$.

To sum up, I have proven that the new sample space can be divided into sets of $k$ points such that in each set there exists at least one point valued above (or at) $\frac{1}{k}$, and the probability of choosing such a point is at least $\frac{1}{k}$.