Krdytkyu

If $0<a_1le a_2le cdotsle a_n$, then the following inequality holds：

$$frac{1}{2n^2a_n} {sum_{1le i < jle n}^{} {(a_i-a_j)^2}}le frac{a_1+a_2+cdots + a_n}{n}-sqrt [n]{a_1 a_2 cdots a_n }{le frac{1}{2n^2a_1}sum_{1le i < jle n}^{} {(a_i-a_j)^2}}$$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 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?

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
$$S=frac{a_1+cdots+a_n}n-sqrt[n]{a_1cdots a_n}-frac1{2n^2a_n}sum_{1le i<jle n}(a_i-a_j)^2$$
and
$$T=frac{a_1+cdots+a_n}n-sqrt[n]{a_1cdots a_n}-frac1{2n^2a_1}sum_{1le i<jle n}(a_i-a_j)^2$$
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 $0le a_1leldots le a_n$, $sum_{i=1}^na_i=1$ since $a_nge1/n>0$ on this set. This set is compact, and $S$ is smooth, so $S$ attains its minimum. On the other hand, $Tto-infty$ as $a_1to0^+$, thus the supremum of $T$ on $0<a_1leldots le a_n$, $sum_{i=1}^na_i=1$ is equal to the supremum of $T$ on $epsilonle a_1leldots le a_n$, $sum_{i=1}^na_i=1$ 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 $a_i<sqrt[n]{a_1cdots a_n}$ 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 $a_i'+a_{i+1}'=a_i+a_{i+1}$, $a_1leldotsle a_{i-1}le a_i'le a_{i+1}'le a_{i+2}leldotsle a_n$, 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 $a_ine0$, we obtain
$$f'(0)=frac{a_{i+1}-a_i}nleft(frac1{a_n}-frac{sqrt[n]{a_1cdots a_n}}{a_ia_{i+1}}right)$$
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 $a_ine0$, it reduces to prove $a_ia_{i+1}<a_nsqrt[n]{a_1cdots a_n}$. But $a_i<sqrt[n]{a_1cdots a_n}$ and $a_{i+1}le a_n$. This proves our lemma.

Lemma 2. Suppose $a_j>sqrt[n]{a_1cdots a_n}$ 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 $a_{j-1}'+a_j'=a_{j-1}+a_j$, $a_1leldotsle a_{j-2}
le a_{j-1}'le a_j'le a_{j+1}leldotsle a_n$, 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 $a_1gesqrt[n]{a_1cdots a_n}$ implies $a_1=a_2=cdots=a_n$; so is $a_nlesqrt[n]{a_1cdots a_n}$. 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.