Lower bounding the probability that a zero-mean sequence of random variables stays positive Assume that $X_n$ is a sequence of a zero-mean and unit variance random variables (and maybe having density w.r.t. to Lebesgue). Can we conclude that $ P(X_n \in [0,R_n]) $ is bounded away from zero eventually, say $$\liminf_{n \to \infty} P(X_n \in [0,R_n]) > 0$$ assuming that $R_n \to \infty$. Intuitively, $X_n$ should put some nonvanishing amount of mass on the positive real line because of the zero-mean assumption and the condition of unit variance should prevent that mass from escaping to infinity (?)
Here is a partial argument: We have
\begin{align*}
P(X_n \in [0,R_n]) &= 1 - P(X_n > R_n) - P(X_n < 0) \\
&\ge 1 - P(|X_n| > R_n) - P(X_n < 0) \\
&\ge 1 - \frac{E X_n^2}{R_n^2} - P(X_n < 0) \\
&= 1 - o(1) - P(X_n < 0)
\end{align*}
Thus, the problem reduces to arguing that $P(X_n < 0)$ stays away from 1.
 A: A standard technique to lower-bound $\mathbb P(0 < X < r)$ as a function of $\mathbb E[|X|^3]$ subject to $\mathbb E[X] = 0$ and $\mathbb E[X^2] = 1$
is to consider a suitable linear combination of $|X|^3$, $X$, $X^2$ and $1$ that is negative on $(0,r)$ and nonnegative outside that interval and has a negative expected value.  In this case try
$$ f(x) = |x|^3 - b r x^2  - (1-b) r^2 x$$ 
For $x \ge 0$ we have $f(x) = x (x-r) (x-(b-1)r)$, so this satisfies the sign requirements there if $b < 1$.
$f(x) > 0$ for $x < 0$ if $b^2 + 4 b - 4 < 0$, which is true if $0 \le b \le 2\sqrt{2}-2$.
We have $\mathbb E[f(X)] = \mathbb E[|X|^3] - b r$, which we want to be negative.  The minimum value of $f(x)$ on $[0,r]$ turns out to be
$$ v = - \dfrac{r^3}{27} \left( 2 (b^2 - 3 b + 3)^{3/2} + 2 b^3 - 9 b^2 + 9 b\right)$$
The conclusion is then that if $\mathbb E[|X|^3] < b r$ where $0 < b \le 2\sqrt{2}-2$, 
$$ \mathbb P(0 < X < r) \ge \frac{ br - \mathbb E[|X|^3]}{-v} $$ 
Taking $b = 2\sqrt{2}-2$, 
we get $$\mathbb P(0 < X < t E[|X|^3]) \ge c \dfrac{t-(1+\sqrt{2})/2}{t^3 \mathbb E[|X|^3]^2}$$
for $t > (1+\sqrt{2})/2$, where $c \approx 4.43035992$.
A: Here's a proof if a third moment condition is satisfied. 
Suppose that $\mathbb EX=0$, $\mathbb EX^2=1$ and $\mathbb E|X|^3\le K$. 
Then let $Z=|X|$. We use Cauchy-Schwarz: $Z^2=Z^{1/2}Z^{3/2}$, so that $(\mathbb EZ^2)^2\le \mathbb EZ\cdot \mathbb EZ^3$. This gives $\mathbb EZ\ge \frac 1K$. Hence $\mathbb EX\mathbf 1_{X>0}\ge \frac{1}{2K}$. 
Also $\mathbb EX^2\mathbf 1_{X>0}\le 1$. Now $\mathbb EX\mathbf 1_{X>4K}\le (1/4K)\mathbb EX^2\mathbf 1_{X>4K}\le 1/(4K)$, so that $\mathbb EX\mathbf 1_{X\in [0,4K]}\ge 1/(4K)$ and $\mathbb P(X\in [0,4K])\ge 1/(16K^2)$. 
A: Filling in the details for Anthony's argument: 
Assume that $\mathbb E |X|^3 \le c$ for numerical constant $c > 0$.
Let $X^+ = X \mathbf 1_{X > 0}$ and $X^- = (-X) \mathbf 1_{X < 0}$. Then, $X = X^+ - X^-$. Let $\mu = \mathbb E X^+ = \mathbb E X^-$. Then, by Anthony's argument $2 \mu = \mathbb E |X| \ge 1/c$.
For $a \le \mu$, we have  $\mathbb E X^+ \le a + \mathbb{E} X^+ \mathbb1_{X^+ > a} \le a + [\mathbb E (X^+)^2]^{1/2} [\mathbb P(X^+ > a)]^{1/2} $ or
\begin{align}
 \mathbb P(X^+ > a) \ge \frac{(\mu-a)^2}{\mathbb E (X^+)^2} \ge (\mu-a)^2 \quad (*)
\end{align}
using $\mathbb E (X^+)^2 \le \mathbb E X^2 \le 1$. Take $a = \frac1{4c}$ so that $a \le \frac1{2c} \le \mu$. Then, $\mathbb P(X^+ > \frac1{4c}) \ge \frac1{16c^2}$. Same bound holds for $X^-$ by symmetry. 
We can conclude that $X$ does not belong to $[-\frac{1}{4c},\frac{1}{4c}]$ with probability at least $1 -\frac{1}{8c^2}$.
EDIT: Apparently $(*)$ is called Paley-Zygmund inequality if one takes $a = \theta \mu$ for $\theta \in [0,1]$.
