Bounding sum of quartic deviations from sample mean [Cross-posted here with no answers for a few days]
I came - to the very best of my knowledge from reading the source - across the following statement in The Jackknife and Bootstrap, Shao and Tu, p. 87:
$$\sum_i(X_i-\bar X)^4\leq16\sum_i(X_i-\mu)^4,$$
where $\bar X$ and $\mu$ are the sample mean and expected value of the $X_i$, respectively.
The statement appears in the broader context of their Theorem 3.8, which of course has several conditions. However, since the result seems to be algebraical, I suppose none of these matter for the present question.
Even without the term 16, the statement is well-known (for any $a$, not only $\mu$) when we replace 4 by 2.
My hunch is that the proof uses Pascal's triangle, as there are 16 terms in the r.h.s. when adding and subtracting $\mu$ and multiplying out:
$$\sum_i(X_i-\bar X)^4=\sum_i(X_i-\mu)^4+4(X_i-\mu)^3(\mu-\bar X)+4(X_i-\mu)(\mu-\bar X)^3+6(X_i-\mu)^2(\mu-\bar X)^2+(\mu-\bar X)^4$$
My attempts at bounding the terms on the r.h.s. (other than the first one, which is already in the "right" format), including Hölder's inequality, however have not led anywhere useful.
 A: So, I believe we have a proof - actually provided by my PhD Student Stephan Hetzenecker, who is however not here, so I'll post it in his name!
W.l.o.g., let $\mu = 0$ (otherwise define $Z_i := X_i - \mu$ and show $\sum_i(Z_i-\bar Z)^4 \leq 16 \sum_i Z_i^4  $).
Using the binomial theorem, we obtain
\begin{align} \tag{1}\label{eq:sum_binom}
\sum_i(X_i-\bar X)^4= &\sum_i \left( X_i^4- 4X_i^3\bar X
       -4X_i\bar X^3
       +6X_i^2\bar X^2    
       +{\bar X}^4\right).
\end{align}
Let $p \geq 1$. Using Jensen's inequality, we get
\begin{align} \tag{2}\label{eq:applyJensen_p}
       \bar X^p := \left(\frac{1}{n} \sum_i X_i \right)^p  \leq  \frac{1}{n} \sum_i X_i^p.
\end{align}
Hence,  applying \eqref{eq:applyJensen_p}  $n$ times for $p=4$,
\begin{align} \tag{3}\label{eq:bound_barx4}
   \sum_i    \bar X^4 = \sum_i \left(\frac{1}{n} \sum_j X_j \right)^4  \leq \sum_i  \frac{1}{n} \sum_j X_j^4 =  \sum_j X_j^4 .
\end{align}
Similarly, using Hölder's inequality with $p=q=2$ and using  \eqref{eq:bound_barx4},
\begin{align} \tag{4}\label{eq:bound_barx2}
   \sum_i X_i^2   \bar X^2   &\leq \left(\sum_i X_i^4 \right)^{1/2} \left(\sum_i \bar X^4 \right)^{1/2}\\&  \leq  \left(\sum_i X_i^4 \right)^{1/2} \left( \sum_i X_i^4 \right)^{1/2}  \\&=  \sum_j X_j^4 .
\end{align}
Furthermore,  using \eqref{eq:applyJensen_p} again
\begin{align} \tag{5}\label{eq:bound_barx3}
   - \sum_i X_i\bar X^3 &= - n \bar X^4 \leq n \bar X^4 \\&\leq  n \left(\frac{1}{n} \sum_j X_j^4 \right) \\&= \sum_j X_j^4
\end{align}
Using Hölder's inequality with $p=4/3$ and $q=4$ and using  \eqref{eq:bound_barx4},
\begin{align} \tag{6}\label{eq:bound_barx}
   -\sum_i  X_i^3\bar X &\leq \sum_i  \vert X_i^3\bar X \vert \\&\leq \left( \sum_i   X_i^4 \right)^{3/4} \left( \sum_i  \bar X^4 \right)^{1/4} \\&\leq \sum_i   X_i^4
\end{align}
Combining \eqref{eq:sum_binom}, \eqref{eq:bound_barx4}, \eqref{eq:bound_barx2}, \eqref{eq:bound_barx3}, and \eqref{eq:bound_barx}, we obtain
\begin{align*}
\sum_i(X_i-\bar X)^4 \leq 16 \sum_i X_i^4,
\end{align*}
which was to be shown.
A: Let $A = 16\sum\limits_{i=1}^{n}(X_i-\mu)^4-\sum\limits_i(X_i-\bar X)^4$
$=\sum\limits_{i=1}^{n}\left(4(X_i-\mu)^2+(X_i-\bar X)^2\right)\left(2(X_i-\mu)+(X_i-\bar X)\right)\left(2(X_i-\mu)-(X_i-\bar X)\right)$
$=\sum\limits_{i=1}^{n}\left(4a_i^2+b_i^2\right)\left(2a_i+b_i\right)\left(2a_i-b_i\right)$, by letting $X_i-\mu=a_i$ and $X_i-\bar{X}=b_i$
Note that $\sum\limits_{i=1}^{n} b_i = \sum\limits_{i=1}^{n}X_i - n\bar{X}= n\bar{X}- n\bar{X} = 0$
Now, notice if $X_i$s are ordered in an increasing sequence, i.e., w.l.o.g. let $X_1\leq X_2\leq\ldots\leq X_n$, we shall have $a_1\leq a_2\leq\ldots \leq a_n$ and $b_1 \leq b_2 \leq \ldots b_n$, since $\mu, \bar{X}$ are constants.
$\implies 4a_1^2+b_1^2\leq4a_2^2+b_2^2\leq\ldots\leq 4a_n^2+b_n^2$ and $2a_1+b_1\leq 2a_2+b_2\leq\ldots 2a_n+b_n$
$\implies (4a_1^2+b_1^2)(2a_1+b_1)\leq(4a_2^2+b_2^2)(2a_2+b_2)\leq\ldots\leq (4a_n^2+b_n^2)(2a_n+b_n)$,  since $4a_i^2+b_i^2 \geq 0$, being sum of squares of real numbers
Similarly, $2a_i-b_i=X_i-2\mu+\bar{X}$
$\implies 2a_1-b_1\leq 2a_2-b_2\leq\ldots\leq 2a_n-b_n$, since $X_1\leq X_2\ldots\leq X_n$
Now, let's use the following inequality from Chebyshev:

Now, applying Chebysev's inequality twice, we have,
$\frac{A}{n}=\frac{1}{n}
\sum\limits_{i=1}^{n}\left(4a_i^2+b_i^2\right)\left(2a_i+b_i\right)\left(2a_i-b_i\right)\geq \left(\frac{1}{n} \sum\limits_{i=1}^{n}(4a_i^2+b_i^2)(2a_i+b_i)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i-b_i)\right)$
$\geq \left(\left(\frac{1}{n}\sum\limits_{i=1}^{n}(4a_i^2+b_i^2)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i+b_i)\right)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i-b_i)\right)$
$= \left(\frac{1}{n} \sum\limits_{i=1}^{n}(4a_i^2+b_i^2)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i)\right)$, since $\sum\limits_{i=1}^{n}b_i=0$
$\geq \left(\frac{1}{n} \sum\limits_{i=1}^{n}(4a_i^2+b_i^2)\right)\left(\frac{1}{n} \sum\limits_{i=1}^{n}(2a_i)\right)^2 \geq 0$, being product of sum of squares of
real numbers
$\implies A = 16\sum\limits_{i=1}^{n}(X_i-\mu)^4-\sum\limits_i(X_i-\bar X)^4\geq 0$
$\implies \sum\limits_i(X_i-\bar X)^4 \leq 16\sum\limits_{i=1}^{n}(X_i-\mu)^4$
