$\mathbb{E}[X^4]=1$, $X,Y$ iid, what's the best upper bound of $\mathbb{E}[(X-Y)^4]$? Let $X,Y$ be i.i.d. random variables, $\mathbb{E}[X^4]=1$, what's the best upper bound for $\mathbb{E}[(X-Y)^4]$ ?


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*A trivial upper bound is $16$, since $(X-Y)^4 \leq 8 (X^4+Y^4)$ then take expectation on both sides. However, equality cannot be achieved.

*My guess of the best upper bound will be $8$, achieved when $X$ is uniform at random from $\{-1, +1\}$.
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Your guess is correct. Indeed, it is well known (see e.g. Bertsimas--Popesku, page 781) that real numbers $m_0=1,m_1,\dots,m_{2\ell}$ are the moments of orders $0,1,\dots,2\ell$ of a real-valued random variable $X$ iff the matrix $M:=(m_{i+j})_{i,j=0}^\ell$ is nonnegative-definite, that is, iff all the principal minors of $M$ are $\ge0$; here $\ell$ is a natural number; in our case, $\ell=2$. 
Also, given $\E X^4=1$, we have 
\begin{equation}
 \E(X-Y)^4=2-8m_3m_1+6m_2^2. 
\end{equation}
Thus, the problem is a simple problem of real algebraic geometry, which can be solved algorithmically. Using the Mathematica command Maximize[], we get the result:


Added: Here is an elementary solution, without using Mathematica: Since $m_4=1$, the condition $\det M\ge0$ implies
\begin{equation}
 2m_3m_1\ge m_2^2-m_4+(m_3^2+m_4m_1^4)/m_2\ge m_2^2-1,
\end{equation}
whence
\begin{equation}
 \E(X-Y)^4=2-8m_3m_1+6m_2^2\le2-4(m_2^2-1)+6m_2^2=2m_2^2+6\le8, 
\end{equation}
since $m_2^2\le m_4=1$. The equality in the inequality in question  is attained only if $m_2^2=m_4=1$ and $2m_3m_1=m_2^2-1=0$, that is, only if $\PP(X=1)=\PP(X=-1)=1/2$. 
