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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 not conclude that $X$ belongs 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 $\theta \in [0,1]$.

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 $\theta \in [0,1]$.

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 concluded that $X$ belongs 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 $\theta \in [0,1]$.

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 not conclude that $X$ belongs 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 $\theta \in [0,1]$.

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 $$ \mathbb P(X^+ > a) \ge \frac{(\mu-a)^2}{\mathbb E (X^+)^2} \ge (\mu-a)^2 $$ 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 concluded that $X$ belongs to $[-\frac{1}{4c},\frac{1}{4c}]$ with probability at least $1 -\frac{1}{8c^2}$.

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 concluded that $X$ belongs 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 $\theta \in [0,1]$.