On Arthur Engel Book, First Chapter : Invariance

Question

I started reading Arthur Engel Book on problem solving strategies, and as a secondary school student I still lack some of the mathematical notions needed for some of the problems presented in the book. My question concerns the first chapter of invariance, problem E6 where an algorithm is considered as follow : $x_{n+1} = \frac{x_n+y_n}{2} ; y_{n+1} = \sqrt{x_{n+1}y_n}$ with $0<x_0<y_0$ according to the writer one can prove from the AM-GM inequality that $$x_n < y_n\Rightarrow x_{n+1} < y_{n+1}, y_{n+1}-x_{n+1} < \frac{y_n-x_n}{4}$$ The rest of the proof is easy and pretty much straightforward , I would like to know how to prove this in a step by step process. Notice : I proved that the inequality $x_n<y_n$ holds for every $n \in \Bbb N$ using recurrence relation , thought it seems from the way the solution was given it could be done with the AM-GM inequality.

Answer 1

First let's visualize AM-GM inequality by an example. Take points $P=2$ and $Q=8$ on the number line. Their AM is point $A=5$. Note that arithmetic mean always lies at the midpoint of segment $PQ$. Their GM is point $G=4$. Geometric mean also lies between $P$ and $Q$, and precedes $A$ (this is the very statement of AM-GM inequality) except for the case when $Q$ coincides with $P$, when $A$ and $G$ also coincide with $P$ and $Q$.

So $x_{n+1}$ lies exactly between $x_n$ and $y_n$ on the number line. $y_{n+1}$ being GM of $x_{n+1}$ and $y_n$, will always lie ahead of $x_{n+1}$ ie, $x_{n+1} \lt y_{n+1}$.

To prove your inequality, $$ y_{n+1} = \sqrt{x_{n+1}y_n} \lt \frac{x_{n+1}+y_{n}}{2} = \frac{(x_{n}+y_{n})/2+y_{n}}{2}$$

This gives $$y_{n+1}-x_{n+1} \lt \frac{(x_{n}+y_{n})/2+y_{n}}{2} - \frac{x_{n}+y_{n}}{2} = \frac{y_{n}-x_{n}}{4}$$