Two sequences defined for all naturals by

[quasar987]

Consider $x_1,y_1 \in \mathbb{R}$ such that $x_1>y_1>0$ and $\{x_n\},\{y_n\}$ the two sequences defined for all naturals by

$$x_{n+1}=\frac{x_n+y_n}{2}, \ \ \ \ \ y_{n+1}=\sqrt{x_n y_n}$$

Show that the sequence $\{y_n\}$ is increasing and as $x_1$ for an upper bound.


I would appreciate some help on this one, I have made no progress.

 

[mathman]

The key to the proof is to show x_n+1>y_n+1. Since they are both positive, this is equivalent to showing the squares exhibit the same inequality. A simple calculation will show this to be true.

x_n+1²-y_n+1²=((x_n-y_n)/2)².

The rest is easy. x_n>x_n+1> ,y_n<y_n+1.

Therefore your desired conclusions hold.

 

[M98Ranger]

Sure, I'd be happy to help! Let's start by looking at the first few terms of the sequences:

x_1, y_1
x_2, y_2
x_3, y_3
x_4, y_4
...

We can see that x_n and y_n alternate in increasing and decreasing order. This pattern continues for all n, as we can see from the recursive definitions of x_n and y_n.

Now, let's prove that the sequence \{y_n\} is increasing. We can do this by showing that y_{n+1} > y_n for all n. Using the recursive definition of y_n, we have:

y_{n+1} = \sqrt{x_n y_n}
y_n = \sqrt{x_{n-1} y_{n-1}}

So, to prove that y_{n+1} > y_n, we need to show that \sqrt{x_n y_n} > \sqrt{x_{n-1} y_{n-1}}. We can do this by squaring both sides of the inequality:

x_n y_n > x_{n-1} y_{n-1}

Now, we know that x_n > x_{n-1} and y_n > y_{n-1} (since x_n and y_n alternate in increasing and decreasing order). So, we can rewrite the inequality as:

x_n y_n > x_n y_n

This is true, since x_n and y_n are both positive. Therefore, we have shown that y_{n+1} > y_n for all n, which means that the sequence \{y_n\} is increasing.

Next, let's show that x_1 is an upper bound for the sequence \{y_n\}. To do this, we need to show that y_n < x_1 for all n. We can use induction to prove this.

First, the base case: for n = 1, we have y_1 = \sqrt{x_1 y_1} < x_1, since x_1 and y_1 are both positive and x_1 > y_1.

Now, for the induction step, assume that y_k < x_1 for some k. We want to show that y_{k+1} < x_1. Using the recursive definition of y_n, we have:

y