Prove this Kenneth S. Williams inequality

Solution 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.