Induction on an inequality involving sequences

[Benny]

I just can't get the following question. Can someone help me out?

Q. Let a < a_1 < b_1 and define a_{n + 1} = \sqrt {a_n b_n } ,b_{n + 1} = \frac{{a_n + b_n }}{2} .

a) Prove that a_n \le a_{n + 1} \le b_{n + 1} \le b_n for all n.
b) Deduce that the sequences {a_n} and {b_n} both converge.
c) Prove that they have the same limit. (This limit is called the arithmetic-geometric mean of a_1 and b_1)

Here are my attempts

b) I'll do this one first. By the result of part 'a' {a_n} is increasing and bounded above by b_n while {b_n} is a decreasing sequence and bounded below by a_n. So by the monotonic sequence theorem both sequences converge. Not sure about the argument there, I might have missed a few important points.

a) There might ways other than induction to do this but I can't think of any. So I need to start off with a specific case.

For n = 1: I can get a_1 <= a_(2) but not the b_(2) <= b_(1) part 'properly' and I can't get the a_2 <= b_2 bit at all.

 

[HallsofIvy]

Here are a few comments. If 0< a< b then
1) 0< a²< ab so 0< a< \sqrt{ab}
That gives you "a₂< a₁".
2) 0< a+ b< 2b so 0< (a+b)/2< b.
That gives you "b₂< b₁".
Now, assume that a_n \le a_{n + 1} \le b_{n + 1} \le b_n
for some k and use the same calculations as above to show that
a_{n+1} \le a_{n + 2} \le b_{n + 2} \le b_{n+1}

 

[Benny]

Thanks for the help HallsofIvy.