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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

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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.

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