TITLE: Is Cauchy induction used for proofs other than for AM–GM?
QUESTION [23 upvotes]: The proof by Cauchy induction of the arithmetic/geometric-mean inequality is well known. I am looking for a further theorem whose proof is much neater by this method than otherwise.

REPLY [2 votes]: Another inequality can be proved by Cauchy induction :$$\prod_{i=1}^n(x_0+x_i)\geqslant\left(x_0+\prod_{i=1}^nx_i^{1/n}\right)^n\quad\text{for all}\quad  n=1,2,...\;\text{and}\; x_0,...,x_n\geqslant0.$$ Details can be found here. I don't see a different easy way to do it. While this involves a geometric mean, it doesn't look much like the AM–GM inequality, because there is a product on both sides. Perhaps, though, it could be derived from AM–GM.
Edit:$\quad$As I have later learned, it can indeed be derived from AM–GM. First, write this standard inequality with variables $x_1,...,x_n>0$. Now replace each $x_i\,$ ($i=1,...,n$) respectively by $x_0^2/(x_0+x_i)$. In parallel, replace each $x_i$ respectively by $x_0x_i/(x_0+x_i)$. Add the two consequent inequalities, and the result easily simplifies to the inequality above. (The extension to $x_1,...,x_n\geqslant0$ is trivial.)