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:
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:
(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
by mathematical induction.