If $0<a_1\le a_2\le \cdots\le a_n$, then the following inequality holds: $$\frac{1}{2n^2a_n}{\sum_{1\le i < j\le n}^{} {(a_i-a_j)^2}}\le \frac{a_1+a_2+\cdots + a_n}{n}-\sqrt [n]{a_1 a_2 \cdots a_n }{\le \frac{1}{2n^2a_1}\sum_{1\le i < j\le n}^{} {(a_i-a_j)^2}}.$$

This problem was proposed by Kenneth S. Williams, Carleton University, Ottawa in CRUX 247[1977;131] and in CRUX[1978;23,37] it is said enter image description here that there is a nice simple proof but I can't find this G. Szekeres (October 1977 was published by Rennie in JCMN, NO.12) shorter proof. Can help me? Thanks.

Maybe now this inequality have some methods to solve it, such as AM-GM inequality?


1 Answer 1


This is not a hard inequality, so I can't imagine what the "longer proof" could be. Here is a proof that only involves easy techniques.

By homogeneity we may assume $\sum_{i=1}^na_i=1$. Let $$S=\frac{a_1+\cdots+a_n}n-\sqrt[n]{a_1\cdots a_n}-\frac1{2n^2a_n}\sum_{1\le i<j\le n}(a_i-a_j)^2$$ and $$T=\frac{a_1+\cdots+a_n}n-\sqrt[n]{a_1\cdots a_n}-\frac1{2n^2a_1}\sum_{1\le i<j\le n}(a_i-a_j)^2.$$ Obviously, if $a_1=\cdots=a_n=1/n$, we have $S=T=0$. It suffices to prove that $S$ (resp. $T$) attains its minimum (resp. maximum) at $a_1=\cdots=a_n=1/n$. We first show that the minimum of $S$ and the maximum of $T$ are attainable. In fact, the domain of $S$ can be extended to non-negative reals $0\le a_1\le\dots \le a_n$, $\sum_{i=1}^na_i=1$ since $a_n\ge1/n>0$ on this set. This set is compact, and $S$ is smooth, so $S$ attains its minimum. On the other hand, $T\to-\infty$ as $a_1\to0^+$, thus the supremum of $T$ on $0<a_1\le\dots \le a_n$, $\sum_{i=1}^na_i=1$ is equal to the supremum of $T$ on $\epsilon\le a_1\le\dots \le a_n$, $\sum_{i=1}^na_i=1$ for some $\epsilon>0$. Again this is a compact set and $T$ attains its maximum on it.

The key point of the proof is the following two lemmas:

Lemma 1. Suppose $a_i<\sqrt[n]{a_1\cdots a_n}$ for some $i$. In this case, we may choose $i$ to be the maximum among all such $i$. Then there exists a pair $a_i',a_{i+1}'$ such that $a_i'+a_{i+1}'=a_i+a_{i+1}$, $a_1\le\dots\le a_{i-1}\le a_i'\le a_{i+1}'\le a_{i+2}\le\dots\le a_n$, and \begin{align}S(a_1,\ldots,a_i',a_{i+1}',\ldots,a_n)<S(a_1,\ldots,a_i,a_{i+1},\ldots,a_n).\end{align}

Proof. By the maximality of $i$ we have $a_i<a_{i+1}$. Let $f(\lambda)=S(a_1,\ldots,a_i+\lambda,a_{i+1}-\lambda,\ldots,a_n)$. Taking derivatives if $a_i\ne0$, we obtain $$f'(0)=\frac{a_{i+1}-a_i}n\left(\frac1{a_n}-\frac{\sqrt[n]{a_1\cdots a_n}}{a_ia_{i+1}}\right)$$ We show that $f'(0)<0$, hence for some sufficiently small $\lambda>0$ we may take $a_i'=a_i+\lambda$ and $a_{i+1}'=a_{i+1}-\lambda$ which would strictly decrease $S$ as is required. In light of this, we may admit the case $a_i=0$ here where we think $f'(0)=-\infty<0$. For $a_i\ne0$, it reduces to prove $a_ia_{i+1}<a_n\sqrt[n]{a_1\cdots a_n}$. But $a_i<\sqrt[n]{a_1\cdots a_n}$ and $a_{i+1}\le a_n$. This proves our lemma.

Lemma 2. Suppose $a_j>\sqrt[n]{a_1\cdots a_n}$ for some $j>1$. In this case, we may choose $j$ to be the minimum among all such $j$. Then there exists a pair $a_{j-1}',a_j'$ such that $a_{j-1}'+a_j'=a_{j-1}+a_j$, $a_1\le\dots\le a_{j-2}\le a_{j-1}'\le a_j'\le a_{j+1}\le\dots\le a_n$, and $$T(a_1,\ldots,a_{j-1}',a_j',\ldots,a_n)>T(a_1,\ldots,a_{j-1},a_j,\ldots,a_n).$$

Proof. Completely similar to the proof of Lemma 1.

Finally, we note that $a_1\ge\sqrt[n]{a_1\cdots a_n}$ implies $a_1=a_2=\cdots=a_n$; so is $a_n\le\sqrt[n]{a_1\cdots a_n}$. If the maximum (resp. minimum) of $S$ (resp. $T$) is not attained at $a_1=\cdots=a_n=1/n$, Lemma 1(resp. Lemma 2) would immediately yield a contradiction. This completes our proof.


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