Bound third moment and Fourth moment 1) I have seen it mentioned that if X is a zero mean random variable then $\frac{E[X^4]}{\sigma^4} \geq \left\{\frac{E[X^3]}{\sigma^3}\right\}^2 + 1$ , where $E[X^2]=\sigma^2$ .
Is this true, if yes ,what is the proof for this?
(I am able to prove that $\frac{E[X^4]}{\sigma^4} \geq \left\{\frac{E[X^3]}{\sigma^3} \right\}^2 $)
2) Similar to the above statement, is it possible to find an upper bound or lower bound for the third moment $E[X^3]$ in terms of the second moment $E[X^2]$ , for a zero mean $X$?
I had tried using Holder's and Jensen's inequality to prove these, but it didn't give me the right direction for the bounds. I got bounds like $E[X^3] \leq \sqrt{E[X^2]\cdot E[X^4]} )$. This gives me that $E[X^4] \geq \{E[X^3]/\sigma\}^2$
 A: This is a partial answer to some of the questions. It gives details on my above comments. (Answering will also help me find this question again later if I want.)
On upper/lower bounds for $E[X^3]$
If $E[X]=0$ and $E[X^2]$ is finite, this places no upper or lower bound on $E[X^3]$. Indeed, define $X$ with  PDF 
$$ f_X(x) = (1/2)\delta(x+1/2) + 1\{x\geq 0\} \frac{(3/2)}{(x+1)^4} \quad \forall x \in \mathbb{R}$$
where $1\{x \geq 0\}$ is an indicator function that is 1 if $x\geq 0$ (and zero else); $\delta(x+1/2)$ is a unit impulse at $x=-1/2$. So $P[X=-1/2]=P[X\geq 0]=1/2$ and indeed $\int_{-\infty}^{\infty} f_X(x)dx = 1$. Also, 
$$ E[X]=0, E[X^2] = 5/8, E[X^3] = \infty $$
Fix $a> 0$.  Defining $Z=aX$ gives $E[Z]=0$, $E[Z^3]=\infty$, and $E[Z^2]$ can be any positive value we like (by scalaing $a>0$). 
Likewise if we define $Y=-X$, then we see $E[Y^3]=-\infty$ even though $E[Y]=0$ and $E[Y^2]$ is finite. 
On the conjecture $\frac{E[X^4]}{\sigma^4} \geq 1 + \frac{E[X^3]^2}{\sigma^6}$ for all $X$ such that $E[X]=0$
(Assuming $\sigma^2=Var(X)>0$.) As in my comments above, this is true in the special case $E[X^3]=0$, since it reduces to $E[X^4] \geq E[X^2]^2$ (which is true by Jensen's inequality). 
In the general case $E[X^3]\neq 0$, by defining $Y=X/\sigma$, we see the desired inequality is true if and only if $$E[Y^4] \geq 1 + E[Y^3]^2$$ 
for all random variables $Y$ with $E[Y]=0$, $E[Y^2]=1$. 
On tightness of $E[X^4] \geq 1 + E[X^2]^2$
As in my comments above, fix $m\geq 0$.  We can design a random variable $X$ that satisfies $E[X]=0, E[X^2]=1, E[X^3]=m$, and that satisfies $E[X^4] = 1 + E[X^2]^2$.  This is achieved by: 
\begin{align}
X &= \left\{ \begin{array}{ll}
a &\mbox{ with prob $1/(1+a^2)$} \\
(-1/a)  & \mbox{ with prob $a^2/(1+a^2)$} 
\end{array}
\right. \\
a &= (1/2)\left(m + \sqrt{m^2+4}\right)
\end{align}
Indeed, with $m\geq 0$ we see that $a \geq 1$ and 
$$ E[X] = \frac{(a)(1)}{1+a^2} + \frac{(-1/a)(a^2)}{1+a^2} = 0$$
$$ E[X^2] = \frac{(a)^2(1)}{1+a^2} + \frac{(-1/a)^2(a^2)}{1+a^2} = 1$$
$$ E[X^3] = \frac{(a)^3(1)}{1+a^2} + \frac{(-1/a)^3(a^2)}{1+a^2} = \frac{a^2-1}{a} = m $$
$$ E[X^4] = \frac{(a)^4(1)}{1+a^2} + \frac{(-1/a)^4(a^2)}{1+a^2} =1 + \underbrace{\frac{(a^2-1)^2}{a^2}}_{E[X^3]^2}$$
Scaling $X$ by $\sigma$ gives cases of zero-mean variables with $\frac{E[X^4]}{\sigma^4} = 1 + \frac{E[X^3]^2}{\sigma^6}$. 
Defining $Y=-X$ gives $E[Y]=0, E[Y^2]=1, E[Y^3]=-m$ and $E[Y^4] = 1 + E[Y^2]^2$.
