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