Let $X$ be a real-valued random variable with $X \geq 0$ and $\mathbb E X >0$.
I would like to bound $\mathbb P(X >0)$ from below using information about the first few moments of the variable.


From Cauchy-Schwarz inequality we know $\mathbb P(X>0) \geq \frac{(\mathbb E X)^2}{\mathbb E X^2 }$. Unfortunately, this is not a good enough bound (for instance, if you apply it to $X$ being the absolute value of a standard Gaussian you get $\mathbb P(X>0) \geq \frac{2}{\pi}=0.636..$). 


Is there a refined lower bound?



non-negative random variable
Let $f(x) = \sum\_{j=0}^d a\_j x^j$ be any polynomial of degree $d$ such that
$f(0) = 0$ and $f(x) \le 1$ for all $x \ge 0$. Then 
$$P(X > 0) \ge E[f(X)] = \sum\_{j=0}^d a\_j E[X^j]$$
Your Cauchy-Schwarz bound is the case $f(x) = 1 - (x - c)^2/c^2 = 2 x/c - x^2/c^2$ where $c = E[X^2]/E[X]$. 


If you want a bound that scales properly (so the bound for $X$ is the same as the bound for $kX$ for any $k > 0$), you can take $a\_j = b\_j/E[X]^j$. 


These bounds are best possible in the sense that they are exact for any $X$ whose distribution is concentrated on $0$ and the points where $f(x) = 1$. 


If $X$ is a random variable supported on $[0, L]$, you can take a polynomial $f$ such that $f(0) = 0$,
$0 \le f(x) \le 1$ for $0 \le x \le 1$, and $f(x) \ge 1-\epsilon$ for $\delta \le x \le L$, and then $E[f(X)] \ge (1-\epsilon) P(X \ge \delta)$. So you can 
tailor the estimate to be as close to $P(X > 0)$ as you want for this particular distribution.


