Show that $|y_{n+1} - x_{n+1}| \le \frac{|b-a|}{4^n}$ for $x_{n+1} ={1\over 2}(x_n + y_n)$ and $y_{n+1} = \sqrt{{1\over 2}(x_n^2 + y_n^2)}$

Question

This question is the last part of the problem statement which was not included in this question.

Let: $$ \begin{cases} x_{n+1} = {1\over 2}(x_n+y_n)\\ y_{n+1} = \sqrt{{1\over 2}(x_n^2 + y_n^2)} \\ x_1 = a > 0\\ y_1 = b > 0 \\ n\in \mathbb N \end{cases} $$ Prove that: $$ |y_{n+1} - x_{n+1}| \le \frac{|b-a|}{4^n} $$

I think the author is expecting the properties from the linked question to be utilized, but i don't see how.

I've started with the following: $$ y_{n+1} - x_{n+1} = \sqrt{{1\over 2}(x_n^2 + y_n^2)} -{1 \over 2} (x_n + y_n) = \\ = \frac{{1\over 2}(x_n^2 + y_n^2) -{1 \over 4} (x_n + y_n)^2}{\sqrt{{1\over 2}(x_n^2 + y_n^2)} +{1 \over 2} (x_n + y_n)} = \frac{{1 \over 4}(x_n - y_n)^2}{\sqrt{{1\over 2}(x_n^2 + y_n^2)} +{1 \over 2} (x_n + y_n)} = \\ = \frac{(x_n - y_n)^2}{4\sqrt{{1\over 2}(x_n^2 + y_n^2)}+2(x_n + y_n)} $$

But i do not see how to proceed from here. I guess I should somehow use that:

$$ \begin{cases} y_n > x_n \\ x_{n+1} > x_n \\ y_{n+1} < y_n \end{cases} $$

Or is it even a wrong starting point? How do i prove what's in the problem statement?

Answer 1

A simple calculation shows that $$ \sqrt{{\frac 12}(x^2 + y^2)} \ge \frac 12 (x+y) $$ for all non-negative real numbers $x, y$ (compare Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality).

Then, continuing with your computation, $$ | y_{n+1} - x_{n+1}| = \frac{\frac 14 (x_n - y_n)^2}{\sqrt{{\frac 12}(x_n^2 + y_n^2)} + \frac 12 (x_n+y_n)} \le \frac{\frac 14 (x_n - y_n)^2}{x_n + y_n} \\ = \frac{|y_n - x_n|}{4} \cdot \frac{|y_n - x_n|}{x_n + y_n} \le \frac{|y_n - x_n|}{4} $$ and the result follows with induction.