Thread: Arithmetic/Geometric Means

I'm supposed to prove that:

If 0<y(1)<x(1) and x(n+1)=(x(n)+y(n))/2 and y(n+1)=sqrt(x(n)y(n)
then 0<y(n)<x(n).

I can't figure out how to get started on this proof. I can see that it should be an induction proof...but I can't figure out how to prove it. Any help would be great.

Originally Posted by dancingshoes63

I'm supposed to prove that:

If 0<y(1)<x(1) and x(n+1)=(x(n)+y(n))/2 and y(n+1)=sqrt(x(n)y(n)
then 0<y(n)<x(n).

I can't figure out how to get started on this proof. I can see that it should be an induction proof...but I can't figure out how to prove it. Any help would be great.

First the result to be proven is true for n=1, the base case as its give.

You do know the arithmetic geometric mean inequality?

For any positive numbers a, b:

(a+b)/2 >= sqrt(a.b)

with equality if and only if a=b.

(in this two variable version this is a corrolary of Pythagoras's theorem)

Now applying this you should see that if for some k:

0<y(k)<x(k),

as x(k+1) is the arithmetic mean of y(k) and x(k), and y(k+1) is
the geometric mean of x(k) and y(k) we have by the arithmetic
geometric mean inequality:

y(k+1)<x(k+1)

(we have strict inequality because y(k)!=x(k)), and y(k+1)>0
as it is the square root of the product of two positive numbers.

That is sufficient, together with the base case, to conclude that

0<y(n)<x(n)

by mathematical induction.

RonL

So to prove the arithmetic/geometric mean inequality, I should look at the pythagorean theorem and go from there?

Originally Posted by dancingshoes63

So to prove the arithmetic/geometric mean inequality, I should look at the pythagorean theorem and go from there?

Just consider a right triangle with sides (a+b)/2, (a-b)/2, sqrt(a.b), and
the fact that the length of the hypotenuse is >= any other side.

RonL

Or you can be more elegant.
You know that,
(sqrt(a)-sqrt(b))^2=>0
Thus,
a-2\sqrt(ab)+b>=0
Thus,
a+b>=2\sqrt{ab}