This question already has an answer here:

(Arithmetic–Geometric Mean).

(a) Explain why $\sqrt{xy}≤ (x+y)/2$ for any two positive real numbers x and y. (The geometric mean is always less than the arithmetic mean.)

(b) Now let $0≤x_1 ≤y_1$ and define $x_{n+1} = \sqrt{x_{n}y_{n}}$ and $y_{n+1} = (x_{n} + y_{n})/2$ . Show $\lim x_n$ and $\lim y_n$ both exist and are equal.

My attempt:

(a) I know that I can show this by using $(\sqrt{x} -\sqrt{y})^2 \geq0$ but I dont know how to illustrate, let alone explain it.

(b) I know that $x_{n+1} \leq y_{n+1}$ for all $n \in \mathbb{N}$. But now I need to know how to show that they are equal.


marked as duplicate by Arnaud D., Gibbs, José Carlos Santos real-analysis Nov 12 '18 at 14:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ For ( b) you might want to show both sequences are bounded and monotone. Then to show they are equal, you use the recursive definition of the $x_n$'s and $y_n$'s. $\endgroup$ – aquilegia.taylor Nov 10 '15 at 3:32

Hints for (b).

First show that the sequence $\{x_n\}$ is always increasing and $\{y_n\}$ is always decreasing. Use this to show that the limits you are considering do exist.

Next, note that $$\eqalign{0 &\le y_{n+1}-x_{n+1}\cr &=\frac{x_n+y_n}2-\sqrt{x_ny_n}\cr &=\frac{y_n-x_n}2 \frac{\sqrt{y_n}-\sqrt{x_n}}{\sqrt{y_n}+\sqrt{x_n}}\cr &<\frac{y_n-x_n}2\cr}$$ and so $y_n-x_n\to0$ as $n\to\infty$.

See if you can finish the argument from here.


Hint for (a):-

Let,$c$ and $d$ e any real numbers.



So,$\frac{c^2+d^2}{2}\geq cd$

Now,let $c=\sqrt a$ and $d= \sqrt b$

Putting this in the equation we have $\frac{a+b}{2}\geq \sqrt {ab}$

(This be extended to $n$ numbers.How?)


Not the answer you're looking for? Browse other questions tagged or ask your own question.