Yet another proof of the inequality between the arithmetic and the geometric means

This is an exposition of the proof of the inequality between arithmetic and geometric means given by A. Pełczyński (1992), “Yet another proof of the inequality between the means”, Annales Societatis Mathematicae Polonae. Seria II. Wiadomości Matematyczne, 29, p. 223–224. The proof might look bizarre, but I can guess some relation with another paper of the author where he proves uniqueness of the John ellipsoid. And as bizarre as it is, and despite the abundance of proofs of this inequality, I found it nice. (The paper is written in Polish, but the formulas allow to understand it.)

For $n\geq 1$, let $a_1,\dots,a_n$ be positive real numbers. Their arithmetic mean is $$ A = \dfrac1n \left(a_1+\dots + a_n\right)$$ while their geometric mean is $$G = \left(a_1\dots a_n\right)^{1/n}.$$ The inequality of the title says $G\leq A,$ with equality if and only if all $a_k$ are equal. By homogeneity, it suffices to prove that $A\geq 1$ if $G=1$, with equality if and only if $a_k=1$ for all $k.$ In other words, we have to prove the following theorem.

Theorem. — If $a_1,\dots,a_n$ are positive real numbers such that $a_1\dots a_n=1$ and $a_1+\dots+a_n\leq n,$ then $a_1=\dots=a_n=1.$

The case $n=1$ is obvious and we argue by induction on $n$.

Lemma. — If $a_1\cdots a_n=1$ and $a_1+\dots+a_n\leq n,$ then $a_1^2+\dots+a_n^2\leq n$

Indeed, we can write $$ a_1^2+\dots+a_n^2 = (a_1+\dots+a_n)^2 - \sum_{i\neq j} a_i a_j \leq n^2 - \sum_{i\neq j} a_i a_j,$$ and we have to give a lower bound for the second term. For given $i\neq j$, the product $a_i a_j$ and the remaining $a_k$, for $k\neq i,j$, are $n-1$ positive real numbers whose product is equal to $1$. By induction, one has $$ n-1 \leq a_i a_j + \sum_{k\neq i,j}a_k.$$ Summing these $n(n-1)$ inequalities, we have $$ n(n-1)^2 \leq \sum_{i\neq j} a_i a_j + \sum_{i\neq j} \sum_{k\neq i,j} a_k.$$ In the second term, every element $a_k$ appears $(n-1)(n-2)$ times, hence $$ n(n-1)^2 \leq \sum_{i\neq j} a_i a_j + (n-1)(n-2) \sum_{k} a_k \leq \sum_{i\neq j} a_i a_j + n(n-1)(n-2), $$ so that $$ \sum_{i\neq j} a_i a_j \geq n(n-1)^2-n(n-1)(n-2)=n(n-1).$$ Finally, we obtain $$a_1^2+\dots+a_n^2 \leq n^2-n(n-1)=n,$$ as claimed.

We can iterate this lemma: if $a_1+\dots+a_n\leq n$, then $$a_1^{2^m}+\dots+a_n^{2^m}\leq n$$ for every integer $m\geq 0$. When $m\to+\infty$, we obtain that $a_k\leq 1$ for every $n$. Since $a_1\dots a_n=1$, we must have $a_1=\dots=a_n=1$, and this concludes the proof.