Cfrgtkky

Let ${x_n}$ and ${y_n}$ be sequences defined by recurrence relations:
$$
begin{cases}
x_{n+1} = {1over 2}(x_n+y_n)\
y_{n+1} = sqrt{{1over 2}(x_n^2 + y_n^2)} \
x_1 = a > 0\
y_1 = b > 0 \
nin mathbb N
end{cases}
$$
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 $x_2$:
$$
x_2 = {1over 2}(a + b)\
y_2 = sqrt{{1over 2}(a^2 + b^2)}
$$

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

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

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

$$
y_n > x_ntag*{$blacksquare$}
$$

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:
$$
frac{x_{n+3}}{x_{n+2}} = frac{{1over 2}(x_{n+2} + y_{n+2})}{{1over 2}(x_{n+1} + y_{n+1})} = frac{x_{n+2} + y_{n+2}}{x_{n+1} + y_{n+1}} stackrel{y_n ge x_n}{ge} frac{2x_{n+2}}{x_{n+1} + y_{n+1}}
$$

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.

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