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Let $\{x_n\}$ and $\{y_n\}$ be sequences defined by recurrence relations: $$ \begin{cases} x_{n+1} = {1\over 2}(x_n+y_n)\\ y_{n+1} = \sqrt{{1\over 2}(x_n^2 + y_n^2)} \\ x_1 = a > 0\\ y_1 = b > 0 \\ n\in \mathbb N \end{cases} $$ Prove that:

  1. $\{\forall n \ge 2: y_n \ge x_n\}$
  2. $\{\forall n \ge 2: x_{n+1} > x_n\}$
  3. $\{\forall n \ge 2: y_n> y_{n+1}\}$

First notice that $x_n > 0$ and $y_n > 0$. We'll need that fact during the proofs.

Statement $(1)$

$\Box$ Check for $y_2$ and $x_2$: $$ x_2 = {1\over 2}(a + b)\\ y_2 = \sqrt{{1\over 2}(a^2 + b^2)} $$

Suppose $y_2 > x_2$: $$ \sqrt{{1\over 2}(a^2 + b^2)} > {1\over 2}(a + b) \iff \\ \iff {1\over 2}(a^2 + b^2) > \left({1\over 2}(a + b)\right)^2 \iff \\ \iff a^2 + b^2> \frac{a^2 + 2ab + b^2}{2} \iff 2a^2 + 2b^2>a^2 + 2ab + b^2 \iff\\ \iff a^2 + b^2>2ab $$ This is true. Suppose $y_{n+1} > x_{n+1}$: $$ \frac{y_{n+1}}{x_{n+1}} = \frac{\sqrt{{1\over 2}(x_n^2 + y_n^2)}}{{1\over 2}(x_n + y_n)} \iff \\ \iff \left(\frac{y_{n+1}}{x_{n+1}}\right)^2 = 2\cdot\frac{(x_n^2 + y_n^2)}{x_n^2 + 2x_ny_n+y_n^2} $$

We need: $$ \frac{(x_n^2 + y_n^2)}{x_n^2 + 2x_ny_n+y_n^2} > {1\over 2} \iff 2x_n^2 + 2y_n^2>x_n^2 +2x_ny_n + y_n^2 \iff \\ \iff x_n^2 + y_n^2 > 2x_ny_n $$

Which yields a true statements for $x_n, y_n >0$. Thus:

$$ y_n > x_n\tag*{$\blacksquare$} $$

Statement $(2)$

$\Box$ I'm skipping the base case for induction, it's very similar to the above and in the end yields:

$$ a^2 + b^2 > 2ab $$

Suppose $x_n < x_{n+1}$, consider the following fraction: $$ \frac{x_{n+3}}{x_{n+2}} = \frac{{1\over 2}(x_{n+2} + y_{n+2})}{{1\over 2}(x_{n+1} + y_{n+1})} = \frac{x_{n+2} + y_{n+2}}{x_{n+1} + y_{n+1}} \stackrel{y_n \ge x_n}{\ge} \frac{2x_{n+2}}{x_{n+1} + y_{n+1}} $$

We want it to be greater than $1$:

$$ \frac{2x_{n+2}}{x_{n+1} + y_{n+1}} > 1 \iff 2x_{n+2} > x_{n+1} + y_{n+1} > 2x_{n+1} \implies x_{n+2} > x_{n+1} $$

Thus: $$ x_{n+2} < x_{n+3} $$

which completes the induction $\tag*{$\blacksquare$}$

Statement $(3)$

This is done similarly to case $(2)$. Once again the base case for $y_2$ and $y_3$ yields $a^2 + b^2 > 2ab$. The assumption here is $y_n > y_{n+1}$. Then we need to show that: $$ \frac{y_{n+3}}{y_{n+2}} < 1 $$

So suppose:

$$ \frac{y_{n+3}}{y_{n+2}} < 1 \iff \frac{x_{n+2}^2 + y_{n+2}^2}{x_{n+1}^2 + y_{n+1}^2} < 1 \iff x_{n+2}^2 + y_{n+2}^2 < x_{n+1}^2 + y_{n+1}^2 \iff \\ \iff x_{n+2}^2 - x_{n+1}^2 < y_{n+1}^2 - y_{n+2}^2 $$

We know by $(2)$ that $x_n$ is increasing, therefore:

$$ 0 < x_{n+2}^2 - x_{n+1}^2 < y_{n+1}^2 - y_{n+2}^2 \iff 0 < y_{n+1}^2 - y_{n+2}^2 $$

Since both $x_n$ and $y_n$ are greater than $0$: $$ y_{n+2}^2 < y_{n+1}^2 \iff y_{n+2} < y_{n+1} $$

Thus $y_n$ is decreasing.

I'm kindly asking to verify my proofs as otherwise i have no one to refer to. Thank you.

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In Statement (2) you showed the fact you where supposing, probably just some typo with the induction. The rest looked fine

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  • $\begingroup$ you are right, i've updated the question. Thanks for spotting. $\endgroup$ – roman Nov 15 '18 at 10:45

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