This should really be a comment, but it just takes too much space to put in the comment box. I was quite puzzled by Ryan's remark that Feige's problem (with just some constant) is hard while it is a three-liner essentially known to Bernstein. I haven't read Feige's paper, to be honest, so I won't be surprised if it finally turns out that it is exactly what is written there. Still, I decided to post in the hope that someone will clarify what's going on here faster than I find Feige's paper.

Step 1: (trivial reformulation)

Let $Y_k=1-X_k$. Then $Y_k\le 1$ and $EY_k=0$. Put $Y=\sum Y_k$. We need to estimate the probability $P$ that $Y\ge -1$.

Step 2: (Bernstein trick). $Ee^{tY}=\prod_k Ee^{tY_k}$.

Now we have to consider 2 cases.

Case 1: $Ee^Y\le 2$. Then $1\le E^{Y/2}\le (1-P)e^{-1/2}+\sqrt{2P}$, and some lower bound for $P$ follows.

Case 2: $Ee^Y>2$. Then we can find $t\in(0,1)$ such that $Ee^{tY}=2$.

Now observe that if $Z\le 1$ is a mean zero random variable, then $Ee^{2Z}\le (Ee^Z)^K$ for some fixed $K$ (the best $K$ in the inequality $F(2z)\le KF(z)$ for the function $F(z)=e^z-1-z$ with $z\in (-\infty,1]$ will certainly work). Applying this observation to each factor in the Bernstein trick, we get $Ee^{2tY}\le 2^K$. Now take $q=2^{-K-1}$, write $$ \frac 12\le E(e^{tY}-qe^{2tY}-1) $$ and note that we take an expectation of a function bounded from above by $\frac 1{4q}$ and negative whenever $Y<0$. So, in this case, we even have a bound on $P(Y>0)$.

This should really be a comment, but it just takes too much space to put in the comment box. I was quite puzzled by Ryan's remark that Feige's problem (with just some constant) is hard while it is a three-liner essentially known to Bernstein. I haven't read Feige's paper, to be honest, so I won't be surprised if it finally turns out that it is exactly what is written there. Still, I decided to post in the hope that someone will clarify what's going on here faster than I find Feige's paper.

Step 1: (trivial reformulation)

Let $Y_k=1-X_k$. Then $Y_k\le 1$ and $EY_k=0$. Put $Y=\sum Y_k$. We need to estimate the probability $P$ that $Y\ge -1$.

Step 2: (Bernstein trick). $Ee^{tY}=\prod_k Ee^{tY_k}$.

Now we have to consider 2 cases.

Case 1: $Ee^Y\le 2$. Then $1\le Ee^{Y/2}\le (1-P)e^{-1/2}+\sqrt{2P}$, and some lower bound for $P$ follows.

Case 2: $Ee^Y>2$. Then we can find $t\in(0,1)$ such that $Ee^{tY}=2$.

Now observe that if $Z\le 1$ is a mean zero random variable, then $Ee^{2Z}\le (Ee^Z)^K$ for some fixed $K$ (the best $K$ in the inequality $F(2z)\le KF(z)$ for the function $F(z)=e^z-1-z$ with $z\in (-\infty,1]$ will certainly work). Applying this observation to each factor in the Bernstein trick, we get $Ee^{2tY}\le 2^K$. Now take $q=2^{-K-1}$, write $$ \frac 12\le E(e^{tY}-qe^{2tY}-1) $$ and note that we take an expectation of a function bounded from above by $\frac 1{4q}$ and negative whenever $Y<0$. So, in this case, we even have a bound on $P(Y>0)$.

In response to Lucia's question

It turns out that no new trickery is required here to get some bound depending on $\alpha$ only: the same old argument of Bernstein works perfectly well in this case too.

After centering, we get mean zero random variables $Y_i$ that are $b_i>0$ with probability $\alpha$ and negative otherwise. Now put $Y=\sum Y_i$ and choose $t$ so that $Ee^{tY}=2$. Note that then $\alpha e^{tb_i}\le 2$ for all $i$, so we still are in the bounded from above setting at that moment (with the bound deteriorating as $\alpha\to 0$). Thus, we do exactly the same with $K=\max_{z\le \log\frac 2\alpha}\frac{F(2z)}{F(z)}$ and get some lower bound (small constant times some power of $\alpha$, apparently, which I have no desire to optimize unless somebody really cares about it) on the probability that $Y>0$.