How prove this $S_{2m}S_{2m+2}\ge\left(1-\frac{1}{m(2m+1)}\right)S^2_{2m+1}$ let $x_{i}\in R,i=1,2,\cdots,n$,and $p_{i}\ge 0,i=1,2,\cdots,n$,such
$$p_{1}+p_{2}+\cdots+p_{n}=1$$
and define
$$S_{k}=\sum_{i=1}^{n}p_{i}x^k_{i}-\left(\sum_{i=1}^{n}p_{i}x_{i}\right)^k$$
show that
$$S_{2m}S_{2m+2}\ge\left(1-\dfrac{1}{m(2m+1)}\right)S^2_{2m+1},m\in N^{+}$$
This problem is my student ask me,and I don't prove it.
when I see this 
$$p_{1}+p_{2}+\cdots+p_{n}=1$$
and define
$$S_{k}=\sum_{i=1}^{n}p_{i}x^k_{i}-\left(\sum_{i=1}^{n}p_{i}x_{i}\right)^k$$
I can consider this probability theory：the Univariate discrete random variable 
$$E(X^k)-(E(X))^k=S_{k}$$
then prove
$$\left(E(X^{2m})-(E(X))^{2m}\right)\left(E(x^{2m+2})-(E(X))^{2m+2}\right)\ge\left(1-\dfrac{1}{m(2m+1)}\right)\left(E(X^{2m+1})-(E(X))^{2m+1}\right)^2$$
But also I can't prove this  ,Thank you
 A: This is American Mathematical Monthly problem 11255 proposed in the
November 2006 issue. The solution is from the April 2008 issue.
Problem:
Proposed by Slavko Simic, Mathematical Institute SANU, Belgrade, Serbia.
Let $n$ be a positive integer, $x_1,\dots,x_n$ be real numbers, and
$p_1,\dots,p_n$ be real numbers summing to 1. For $k\geq1$, let $S_k=\sum_{i=1}^np_ix_i^k-(\sum_{i=1}^n p_ix_i)^k.$
Show that for $m\geq1$, $$S_{2m}S_{2m+2}\geq\left({1-{1\over m(2m+1)}}\right)S_{2m+1}^2.$$
Solution: by O.P. Lossers, Eindhoven University of Technology, Eindhoven, The Netherlands.
We assume that $p_1,\dots, p_n$ are nonnegative real numbers
summing to 1, because if $p=(1,-3,3)$ and $x=(1,2,3)$, then $S_2S_4<(2/3)S^2_3$
and the inequality fails.
Define $f:\mathbb{R}^2\to\mathbb{R}$ by
$$f(x,t)={t^2x^{2m+2}\over (2m+2)(2m+1)}+{2tx^{2m+1}\over 2m(2m+1)}+{x^{2m}\over 2m(2m-1)}.$$
Now ${\partial^2\over\partial x^2}f(x,t)=(tx^m+x^{m-1})^2\geq0$, so $f(x,t)$ is a convex function of $x$ for all $t$.
From Jensen's inequality it then follows that, for all $t$,
$$\sum_{i=1}^np_if(x_i,t)-f\left(\sum_{i=1}^np_ix_i,t\right)\geq0,$$
or equivalently,
$${t^2S_{2m+2}\over (2m+2)(2m+1)}+{2tS_{2m+1}\over 2m(2m+1)}+{S_{2m}\over 2m(2m-1)}\geq0.$$
The expression on the left, considered as a polynomial in $t$, must therefore have a nonpositive discriminant:
$$\left[{2S_{2m+1}\over 2m(2m+1)}\right]^2-4{S_{2m+2}\over (2m+2)(2m+1)}\cdot{S_{2m}\over 2m(2m-1)}\leq0,$$
and this is equivalent to the desired inequality.
