Verify proofs related to monotonicity of $x_{n+1} = {1\over 2}(x_n+y_n)$ and $y_{n+1} = \sqrt{{1\over 2}(x_n^2 + y_n^2)}$

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.

1 Answer

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

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